SHM in Spring-Mass System

Q: What is the time period of a spring-mass system?

The time period of a spring-mass system is T = 2π√(m/k), where m is the mass and k the spring constant. It does not depend on the amplitude or on gravity, so a horizontal, vertical or inclined spring with the same mass and spring has the same period.

Q: What happens to the time period when a spring is cut in half?

Each half has twice the spring constant, because kℓ is constant for a given spring. With the same mass the period becomes T/√2. If the two halves are then joined in parallel, the constant becomes 4k and the period halves.

PhysicsOscillationsFor NEET aspirants

A spring-mass system is the standard model of simple harmonic motion: a block of mass $m$ on a light spring of force constant $k$ oscillates with $T = 2 π m / k$ , whether the spring is horizontal, vertical or on an incline. This page covers SHM in a spring-mass system from Hooke's law to the five-step method for any time period, springs with pulleys, collisions, two-block systems, series and parallel springs and the energy method. Spring-mass questions appear almost every year in JEE Main and NEET.

On this page1Hooke's law2Horizontal & vertical3Five-step method4Pulleys5Sudden changes6Two blocks7Combinations8Energy method

Key Formulas - Quick Reference

★ Must learnSpring-mass period: $T = 2 π \frac{m}{k}$ , $f = \frac{1}{2 π} \frac{k}{m}$ (independent of $A$ and $g$ )
★ Must learnVertical spring: extension at rest $x_{0} = \frac{m g}{k}$ , so $T = 2 π \frac{x _{0}}{g}$
A constant force ( $F_{0}$ , $m g sin θ$ ) shifts the mean position by $F_{0} / k$ but does not change $T$
★ Must learnSeries: $\frac{1}{k _{e q}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$ , $T^{2} = T_{1}^{2} + T_{2}^{2}$ ; parallel: $k_{e q} = k_{1} + k_{2}$ , $\frac{1}{T ^{2}} = \frac{1}{T _{1}^{2}} + \frac{1}{T _{2}^{2}}$
Cut spring: $k ℓ$ = constant; $n$ equal pieces each have constant $nk$
Two blocks on one spring: $T = 2 π μ / k$ , reduced mass $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$
Pulley with spring: if the spring stretches $β x$ when the block moves $x$ , $k_{eff} = β^{2} k$
Fixed pulley of inertia $I$ , radius $R$ : $T = 2 π \frac{m + I / R ^{2}}{k}$
After a sudden change: find the new mean position, then $A^{2} = x^{2} + \frac{v ^{2}}{ω ^{2}}$
Energy method: $E = \frac{1}{2} m_{eff} v^{2} + \frac{1}{2} k_{eff} x^{2} \Rightarrow T = 2 π m_{eff} / k_{eff}$

1. Spring Force and Hooke's Law

A light (massless) spring stretched or compressed by $x$ from its natural length pulls or pushes back with a force proportional to $x$ :

★ Must learnHooke's law for a spring:

F = - k x

k

is the force constant (spring constant), in

N m^{- 1}

: the force needed per metre of extension. A stiffer spring has a larger

k

. The minus sign shows that the force always acts back towards the natural length. The energy stored is

U = \frac{1}{2} k x^{2}

Figure 1: Whether the spring is stretched or compressed, its force

F = - k x

points back to the natural-length position

x = 0

. On a smooth floor that position is the mean position.

Compare with the condition for SHM, $F = - k x$ : a block attached to an ideal spring on a smooth surface is an exact SHM, and the spring constant is directly the SHM constant.

2. Horizontal Spring-Block System

Block of mass $m$ , spring constant $k$ , smooth floor. At the natural length the net force is zero: this is the mean position.
Displace the block by $x$ : $ma = - k x$ , so $a = - \frac{k}{m} x$ .
Compare with $a = - ω^{2} x$ :
$ω = \frac{k}{m}, T = 2 π \frac{m}{k}$

$T$ depends only on $m$ and $k$ : not on the amplitude, and not on $g$ . The same spring-block has the same period on the Moon.
$T \propto m$ : four times the mass doubles the period. $T \propto 1/ k$ : a four times stiffer spring halves it.
$T^{2} = \frac{4 π ^{2}}{k} m$ is a straight line through the origin. Plotting $T^{2}$ against $m$ in the laboratory gives $k$ from the slope.

Figure 2:

T \propto m

(left) curves, but

T^{2} = \frac{4 π ^{2}}{k} m

(right) is a straight line through the origin. Slope of the

T^{2}

m

line gives

k = 4 π^{2} / slope

3. Vertical Spring-Block System

When the block hangs from a vertical spring, gravity stretches the spring before any oscillation starts. The block rests where the spring force balances the weight:

k x_{0} = m g \Rightarrow x_{0} = \frac{m g}{k}

This stretched position, not the natural length, is the mean position. Pull the block a further $x$ down. The spring force is $k (x + x_{0})$ upwards and the weight $m g$ downwards:

F_{net} = m g - k (x + x_{0}) = - k x

Figure 3: A hanging block oscillates about the stretched position

x_{0} = m g / k

, not about the natural length. Measured from there, the net force is

- k x

, so

T = 2 π m / k

The net force is again $- k x$ , so $T = 2 π m / k$ , exactly as for the horizontal spring. Gravity only moves the mean position down by $x_{0}$ .

Exam Trick

Period from the static stretch alone. Since $m / k = x_{0} / g$ ,

T = 2 π \frac{x _{0}}{g}

If a question tells you how far the spring stretches when the block hangs at rest, you do not need $m$ or $k$ separately. A $9.8 cm$ stretch always means $T = 2 π 0.01 \approx 0.63 s$ .

3.1 A constant force never changes the period

If a constant force acts on the block (a steady pull $F_{0}$ on a horizontal spring, or the component $m g sin θ$ on a smooth incline), it only shifts the mean position by $F_{0} / k$ (or $m g sin θ / k$ ). Measured from the new mean position the net force is still $- k x$ , so $T = 2 π m / k$ in every case.

Figure 4: A constant force (

F_{0}

on the left,

m g sin θ

on the right) only shifts the mean position. The period stays

T = 2 π m / k

Horizontal spring

Mean position at the natural length. $T = 2 π m / k$ ; the energy $\frac{1}{2} k A^{2}$ is all spring energy.

Vertical spring

Mean position $x_{0} = m g / k$ below the natural length. Same $T = 2 π m / k$ ; measure $x$ and $A$ from the new mean position.

Key idea

Gravity and any other constant force move the mean position; only

m

and

k

set the period.

4. The Five-Step Method for Any Time Period

Almost every SHM problem, linear or angular, is solved with the same routine:

Find the mean position (stable equilibrium): the net force (or torque) is zero and the potential energy is minimum.
Write the mean-position relation, for example $k x_{0} = m g$ .
Displace the body by a small $x$ (or angle $θ$ ) from the mean position.
Write the net force (or torque) in the displaced position.
Reduce it to $F = - K_{SHM} x$ (or $τ = - K θ$ ) using step 2 or small-angle/binomial approximations. Then $T = 2 π m / K_{SHM}$ (or $2 π I / K$ ).

Why it works: the mean-position relation cancels every constant force, so only the part of the force that changes with $x$ survives. That part is the restoring force.

Figure 5: The five-step routine as a chart. Step 2 cancels every constant force (gravity, a steady push), so only

- K_{SHM} x

survives. With a massive pulley the energy route gives

m_{eff} = m + I / R^{2}

in one line.

5. Springs with Pulleys

With a light string and pulley, the block and the spring need not move by the same amount. Find how far the spring stretches when the block moves $x$ ; that ratio fixes the effective spring constant.

Figure 6: The same spring gives different periods because the pulley changes how far the spring stretches for a block displacement

x

: equal in (i),

x /2

in (ii),

2 x

in (iii).

Arrangement	Spring stretch for block displacement $x$	$K_{SHM}$	Period
(i) String over a fixed pulley, spring to the floor	$x$	$k$	$2 π m / k$
(ii) Pulley hangs from the spring, string tied to the floor	$x /2$	$k /4$	$2 π 4 m / k$
(iii) Block on a movable pulley, string from spring to ceiling	$2 x$	$4 k$	$2 π m /4 k$

Exam Trick

The $β^{2}$ rule. If the spring stretches $β x$ when the block moves $x$ , the spring's energy is $\frac{1}{2} k (β x)^{2} = \frac{1}{2} (β^{2} k) x^{2}$ , so $K_{SHM} = β^{2} k$ . Case (ii): $β = \frac{1}{2}$ gives $k /4$ . Case (iii): $β = 2$ gives $4 k$ . No force analysis needed.

Key idea

With pulleys, ask one question: how far does the spring stretch when the block moves $x$ ? The answer

β x

gives

K = β^{2} k

6. Sudden Changes: Collisions, Added or Removed Mass

When a mass lands on, sticks to, or is removed from an oscillating system, the spring constant stays the same but the mean position and the state of motion change. Use this routine:

Find the new mean position: for a vertical spring it shifts by $Δ m g / k$ .
Find the displacement $x$ from the new mean position and the velocity $v$ just after the change (momentum is conserved in a short collision).
New $ω = k / m_{new}$ ; amplitude from
$A^{2} = x^{2} + \frac{v ^{2}}{ω ^{2}}$
Energy of oscillation $= \frac{1}{2} k A^{2}$ .

Figure 7: Momentum gives

v /2

; the spring is unstretched, so all of

\frac{1}{4} m v^{2}

becomes

\frac{1}{2} k A^{2}

A = v m /2 k

. For

m = 1 kg

v = 2 m s^{- 1}

k = 200 N m^{- 1}

A = 10 cm

T = 2 π 2 m / k = 0.63 s

Figure 8: After the mass lands, it oscillates about the new mean position

m g / k

below the natural length. At first contact it is already

x = m g / k

from that mean position and moving at

2 g h

, so

A^{2} = (m g / k)^{2} + 2 g hm / k

Inelastic collisions lose energy. Use momentum conservation for the collision itself, and energy conservation only for the motion after it.

6.1 Blocks that must stay in contact

A block resting on another, or standing on the floor, can lose contact when the spring force reverses. The largest safe amplitude is the distance from the mean position to the point where contact would be lost.

Figure 9: Contact conditions set the largest safe amplitude. Left: stretch

\leq m g / k

above N.L. plus

2 m g / k

compression at M.P. gives

3 m g / k

. Right: the pair must not rise above N.L.,

2 m g / k

above M.P.

6.2 Stacked blocks held by friction

If a block $m$ sits on a block $M$ attached to the spring, the upper block's acceleration $ω^{2} x$ is supplied only by static friction. Friction needed is largest at the extremes, so slipping starts there first.

Figure 10: Static friction on

m

(towards the mean position) supplies

m ω^{2} x

. It is largest at the extremes, so the pair moves together only while

A \leq μ (M + m) g / k

Quick Recall: tap to check

A block on a horizontal spring is hit by an identical block that sticks. Does the period change?

Yes: the mass doubles, so

T

becomes

2

times.

k

is unchanged.

Which law do you use during a short sticking collision, and which after it?

Momentum conservation during it; energy conservation (SHM) after it.

Where does a stacked upper block start to slip first?

At the extremes, where

ω^{2} x

and so the friction needed is largest.

7. Two-Block Systems and Reduced Mass

Two blocks $m_{1}$ and $m_{2}$ joined by a spring on a smooth floor oscillate about their own equilibrium positions, in opposite directions. No external horizontal force acts, so the centre of mass stays at rest: $m_{1} x_{1} = m_{2} x_{2}$ , where $x_{1}$ and $x_{2}$ are the displacements from the equilibrium positions.

Spring deformation $= x_{1} + x_{2}$ . Force on $m_{1}$ : $- k (x_{1} + x_{2}) = m_{1} a_{1}$ .
Substitute $x_{2} = \frac{m _{1}}{m _{2}} x_{1}$ : $a_{1} = - \frac{k ( m _{1} + m _{2} )}{m _{1} m _{2}} x_{1}$ .
So
$ω^{2} = \frac{k ( m _{1} + m _{2} )}{m _{1} m _{2}} = \frac{k}{μ}, T = 2 π \frac{μ}{k}, μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$

Figure 11: With no external force the centre of mass stays still, so

m_{1} x_{1} = m_{2} x_{2}

. The pair behaves like one reduced mass

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

on the same spring.

Both blocks have the same period; the amplitudes divide inversely as the masses, $m_{1} A_{1} = m_{2} A_{2}$ . If one mass is very large, $μ$ becomes the smaller mass, which is the familiar wall-and-block case.

8. Combination of Springs

8.1 Series combination

Springs joined end to end carry the same tension, and their extensions add: $x = x_{1} + x_{2}$ with $F = k_{1} x_{1} = k_{2} x_{2}$ . Then $\frac{F}{k _{e q}} = \frac{F}{k _{1}} + \frac{F}{k _{2}}$ :

\frac{1}{k _{e q}} = \frac{1}{k _{1}} + \frac{1}{k _{2}}

In series the extension of each spring is inversely proportional to its constant: the softer spring stretches more. For identical springs the extensions are equal.

8.2 Parallel combination

Springs side by side have the same extension, and their forces add: $F = - (k_{1} + k_{2}) x$ :

k_{e q} = k_{1} + k_{2}

Figure 12: Series springs share the force; parallel springs share the stretch. A block between two walls is a parallel case: moving it stretches one spring and compresses the other.

Property	Series	Parallel
Same for both springs	Force (tension)	Extension
$k_{e q}$	$\frac{k _{1} k _{2}}{k _{1} + k _{2}}$ (less than either)	$k_{1} + k_{2}$ (more than either)
Period with the same $m$	$T^{2} = T_{1}^{2} + T_{2}^{2}$	$\frac{1}{T ^{2}} = \frac{1}{T _{1}^{2}} + \frac{1}{T _{2}^{2}}$
$n$ identical springs	$k / n$	$nk$

8.3 Cutting a spring

A spring of length $ℓ$ is like pieces joined in series; each short piece stretches less for the same force, so it is stiffer. For one spring $k ℓ$ = constant: $k_{1} ℓ_{1} = k_{2} ℓ_{2} = k ℓ$ .

Figure 13: For one spring,

k ℓ

= constant. Here

ℓ_{1} = 0.56 ℓ

gives

k_{1} = 1.79 k

and

ℓ_{2} = 0.44 ℓ

gives

k_{2} = 2.27 k

. Cutting into

n

equal pieces gives

nk

each; a ratio

p : q

gives

\frac{( p + q ) k}{p}

and

\frac{( p + q ) k}{q}

Exam Trick

Period combos without $k$ . If the same mass has periods $T_{1}$ and $T_{2}$ on two springs, then in series $T = T_{1}^{2} + T_{2}^{2}$ and in parallel $T = \frac{T _{1} T _{2}}{T _{1}^{2} + T _{2}^{2}}$ . For $3 s$ and $4 s$ : series $5 s$ , parallel $2.4 s$ (Solved Example 15).

Quick Recall: tap to check

A spring is cut into two equal halves. What is the constant of each half?

2 k

(for one spring,

k ℓ

is constant).

Two identical springs are joined in series. What is

k_{e q}

k /2

A block sits between two springs fixed to opposite walls. Series or parallel?

Parallel:

k_{e q} = k_{1} + k_{2}

, because both springs push the block back.

9. Energy Method for the Time Period

When several bodies move together (a block and a rotating pulley, a rolling cylinder on a spring), writing forces is slow. Use energy instead: in SHM the total mechanical energy is constant, so $\frac{d E}{d t} = 0$ .

Find the mean position and the mean-position relation (for example $m g = k x_{0}$ ).
Let the body be displaced by $x$ from the mean position with speed $v$ .
Write the total energy $E$ in the displaced position (kinetic energy of every moving part, spring energy, gravitational energy).
Set $\frac{d E}{d t} = 0$ , put $\frac{d x}{d t} = v$ , $\frac{d v}{d t} = \frac{d ^{2} x}{d t ^{2}}$ and use the mean-position relation. The result has the form $m_{eff} \frac{d ^{2} x}{d t ^{2}} + k_{eff} x = 0$ .
Read off $T = 2 π m_{eff} / k_{eff}$ .

Read $m_{eff}$ and $k_{eff}$ straight from $E$ . Write $E$ in the form $\frac{1}{2} m_{eff} v^{2} + \frac{1}{2} k_{eff} x^{2}$ + constant (the constant and linear terms cancel through the mean-position relation). A pulley of inertia $I$ turning at $v / R$ adds $\frac{1}{2} (I / R^{2}) v^{2}$ , so $m_{eff} = m + I / R^{2}$ .

JEE Advanced

Spring with mass. If the spring's own mass $m_{s}$ is not negligible, its kinetic energy is $\frac{1}{2} \cdot \frac{m _{s}}{3} v^{2}$ (each element moves in proportion to its distance from the fixed end). So

T = 2 π \frac{m + m _{s} /3}{k}

The energy method gives this in one line.

Quick Recall: tap to check

A pulley of moment of inertia

I

and radius

R

turns with the string. What is

m_{eff}

m + I / R^{2}

A spring of mass

m_{s}

holds a block

m

. What mass goes into

T

m + m_{s} /3

Does gravity appear in

k_{eff}

for a vertical spring?

No:

m g

is cancelled by the mean-position relation

k x_{0} = m g

Figure 14: Spring-mass SHM on one screen. Only

m

and

k

(or their effective values) ever set the period.

10. Solved Examples

Solved Example 1

A mass

m

is attached to the free end of a massless spring of constant

k

whose other end is fixed to a rigid support. Find the time period if the mass is displaced slightly by

x

downward.

Solution:

Step 1: the hanging position is the mean position. Step 2: $k x_{0} = m g$ . Step 3: displace by $x$ . Step 4: $F_{net} = m g - k (x + x_{0})$ . Step 5: with $k x_{0} = m g$ , $F_{net} = - k x$ .

Answer: $T = 2 π m / k$ , the same as for a horizontal spring.

Solved Example 2

The string, spring and pulley are light. A string over a fixed pulley has a spring (constant

k

, lower end fixed to the floor) on one side and a block of mass

m

on the other. Find the time period.

Solution:

At equilibrium, extension $x_{0}$ with $k x_{0} = m g$ . Displace the block down by $x$ : the spring stretches by the same $x$ , so $F = k (x + x_{0}) - m g = k x$ towards the mean position.

Answer: $T = 2 π m / k$ (Figure 6, case i).

Solved Example 3

A system has a massless pulley, a spring of constant

k

and a block of mass

m

. Find the period of small vertical oscillations if (a) the pulley hangs from the spring and the string, tied to the floor on one side, carries the block on the other; (b) the block hangs from a movable pulley whose string runs from the spring (fixed above) to the ceiling.

Solution:

(a) At equilibrium $T_{0} = m g$ and the spring holds both strands: $k x_{0} = 2 T_{0} = 2 m g$ . If the block moves down $x$ , the pulley moves down $x /2$ and the spring force rises by $k x /2$ , so the string tension rises by $k x /4$ :

F_{net} = m g - \frac{k}{2} (x_{0} + \frac{x}{2}) = - \frac{k x}{4}

$K_{SHM} = k /4$ , so $T = 2 π 4 m / k$ .

(b) Two strands hold the pulley: $2 T_{0} = m g$ and $T_{0} = k x_{0}$ , so $k x_{0} = m g /2$ . If the block (and pulley) moves down $x$ , the spring stretches $2 x$ : $F_{net} = m g - 2 k (x_{0} + 2 x) = - 4 k x$ . So $T = 2 π m /4 k$ .

Solved Example 4

A block of mass

m

moving with speed

v

on a smooth floor collides with an identical block attached to a spring (constant

k

, other end fixed to a wall) and sticks to it. Find the amplitude of the resulting SHM.

Solution:

The collision is very short, so momentum is conserved: common velocity $= v /2$ . Kinetic energy just after $= \frac{1}{2} (2 m) (\frac{v}{2})^{2} = \frac{1}{4} m v^{2}$ . The spring is unstretched at that moment, so this is the whole energy of oscillation: $\frac{1}{2} k A^{2} = \frac{1}{4} m v^{2}$ .

Answer: $A = v \frac{m}{2 k}$ .

Solved Example 5

Blocks

m_{1}

and

m_{2}

hang together at rest from a spring of constant

k

. Block

m_{1}

is suddenly removed. Find the time period and amplitude of the resulting motion of

m_{2}

Solution:

Initial extension: $\frac{( m _{1} + m _{2} ) g}{k}$ . The new mean position for $m_{2}$ alone is $\frac{m _{2} g}{k}$ below the natural length. At the moment of removal $m_{2}$ is at rest, so that point is an extreme.

Answer: $A = \frac{m _{1} g}{k}$ , $T = 2 π m_{2} / k$ .

Solved Example 6

A block

m_{2}

hangs at rest from a spring of constant

k

. A mass

m_{1}

moving vertically downwards with speed

u

collides with

m_{2}

and sticks to it. Find the energy of oscillation.

Solution:

Momentum: $m_{1} u = (m_{1} + m_{2}) v$ , so $v = \frac{m _{1} u}{m _{1} + m _{2}}$ . The new mean position is $\frac{m _{1} g}{k}$ below the old one, so just after the collision $x = \frac{m _{1} g}{k}$ and $ω^{2} = \frac{k}{m _{1} + m _{2}}$ .

From $v^{2} = ω^{2} (A^{2} - x^{2})$ : $k A^{2} = (m_{1} + m_{2}) v^{2} + k x^{2} = \frac{m _{1}^{2} u ^{2}}{m _{1} + m _{2}} + \frac{m _{1}^{2} g ^{2}}{k}$ .

Answer:

E = \frac{1}{2} k A^{2} = \frac{1}{2} [\frac{m _{1}^{2} u ^{2}}{m _{1} + m _{2}} + \frac{m _{1}^{2} g ^{2}}{k}]

Solved Example 7

A body of mass

m

falls from a height

h

onto the pan of a spring balance (pan and spring massless, constant

k

), sticks to it and oscillates vertically. Find the amplitude and the energy of oscillation.

Solution:

It reaches the pan at the natural length with $v = 2 g h$ . The mean position is $m g / k$ lower, so at that moment $x = m g / k$ , and $ω = k / m$ (Figure 8).

$A^{2} = x^{2} + \frac{v ^{2}}{ω ^{2}} = (\frac{m g}{k})^{2} + \frac{2 g hm}{k}$ , so

A = \frac{m g}{k} 1 + \frac{2 k h}{m g}

Energy of oscillation $= \frac{1}{2} k A^{2} = m g h + \frac{( m g ) ^{2}}{2 k}$ .

Solved Example 8

A body of mass

2 m

is fixed on top of a vertical spring whose lower end is attached to a body of mass

m

resting on the ground. The mass

2 m

performs vertical SHM. Find the maximum amplitude so that

m

does not lift off the ground.

Solution:

At the mean position the spring is compressed by $x_{0} = \frac{2 m g}{k}$ . The lower block lifts when the spring pulls it up with a force of at least $m g$ , that is when the spring is stretched by $x^{'} = \frac{m g}{k}$ . The upper block is then $x_{0} + x^{'}$ above its mean position.

Answer: $A_{m a x} = \frac{2 m g}{k} + \frac{m g}{k} = \frac{3 m g}{k}$ (Figure 9, left).

Solved Example 9

A block of mass

m

rests on another block of the same mass attached to a vertical spring of constant

k

. Find the maximum amplitude for which the blocks stay in contact.

Solution:

Mean position: compression $x_{0} = \frac{2 m g}{k}$ . Above the natural length the spring pulls the lower block down, so it decelerates faster than $g$ while the upper block can decelerate only at $g$ : they separate. Contact is kept as long as the pair does not rise above the natural length.

Answer: $A_{m a x} = x_{0} = \frac{2 m g}{k}$ (Figure 9, right).

Solved Example 10

Two blocks

m_{1}

and

m_{2}

are joined by a spring of natural length

ℓ

and constant

k

on a smooth horizontal surface. The spring is compressed by

x_{0}

and released. Show that the blocks perform SHM and find (a) the time period, (b) the amplitude of each block, (c) the length of the spring as a function of time.

Solution:

(a) As derived in Section 7, $ω^{2} = \frac{k ( m _{1} + m _{2} )}{m _{1} m _{2}}$ , so $T = 2 π \frac{μ}{k}$ with $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ .

(b) $m_{1} A_{1} = m_{2} A_{2}$ , and at release both blocks are at their extremes, so $A_{1} + A_{2} = x_{0}$ . Hence $A_{1} = \frac{m _{2} x _{0}}{m _{1} + m _{2}}$ and $A_{2} = \frac{m _{1} x _{0}}{m _{1} + m _{2}}$ .

(c) Take the equilibrium position of $m_{1}$ as origin: $x_{1} = A_{1} cos ω t$ and $x_{2} = ℓ - A_{2} cos ω t$ . Length $= x_{2} - x_{1} = ℓ - (A_{1} + A_{2}) cos ω t = ℓ - x_{0} cos ω t$ .

Solved Example 11

Find the time period of a mass

m

and the equivalent spring constant in each case: (a) springs

k_{1}

and

k_{2}

joined end to end, and this pair placed side by side with a third spring

k_{3}

, all attached to

m

; (b)

m

placed between springs

k_{1}

and

k_{2}

fixed to opposite walls.

Solution:

(a) $k_{1}$ and $k_{2}$ in series: $\frac{k _{1} k _{2}}{k _{1} + k _{2}}$ . In parallel with $k_{3}$ :

k_{e q} = \frac{k _{1} k _{2}}{k _{1} + k _{2}} + k_{3} = \frac{k _{1} k _{2} + k _{2} k _{3} + k _{1} k _{3}}{k _{1} + k _{2}}, T = 2 π \frac{m ( k _{1} + k _{2} )}{k _{1} k _{2} + k _{2} k _{3} + k _{1} k _{3}}

(b) Displace $m$ by $x$ : one spring is stretched by $x$ and the other compressed by $x$ ; both push it back. So $k_{e q} = k_{1} + k_{2}$ and $T = 2 π \frac{m}{k _{1} + k _{2}}$ .

Solved Example 12

The friction coefficient between two blocks is

μ

and the floor is smooth. Block

m

rests on block

M

, which is attached to a spring of constant

k

. (a) Find the time period for small oscillations. (b) Find the friction force when the displacement is

x

. (c) Find the maximum amplitude for which the upper block does not slip.

Solution:

(a) For small amplitude the blocks move together: $ω = \frac{k}{M + m}$ , $T = 2 π \frac{M + m}{k}$ .

(b) Acceleration $a = - ω^{2} x = - \frac{k x}{M + m}$ . Friction supplies the upper block's force: $∣ f ∣ = \frac{mk ∣ x ∣}{M + m}$ .

(c) Friction needed is largest at the extremes, $\frac{mk A}{M + m}$ , and cannot exceed $μ m g$ : $A_{m a x} = \frac{μ ( M + m ) g}{k}$ (Figure 10).

Solved Example 13

A pulley of radius

R

and moment of inertia

I

hangs from the ceiling. A string over it holds a block

m

on one side and is tied to a spring of constant

k

(fixed to the floor) on the other. The string does not slip. Find the period of vertical oscillation.

Solution:

Mean position: $m g = k x_{0}$ . Displace the block by $x$ with speed $v$ ; the pulley turns at $ω_{p} = v / R$ . Taking the gravitational potential energy as zero at the mean position:

E = \frac{1}{2} m v^{2} + \frac{1}{2} I \frac{v ^{2}}{R ^{2}} + \frac{1}{2} k (x + x_{0})^{2} - m g x

$\frac{d E}{d t} = 0$ : $(m + \frac{I}{R ^{2}}) \frac{d ^{2} x}{d t ^{2}} + k x + (k x_{0} - m g) = 0$ , and the bracket is zero.

Answer: $T = 2 π \frac{m + I / R ^{2}}{k}$ .

Solved Example 14

A block hung from a spring stretches it by

9.8 cm

. Find the period of its vertical oscillations. Take

g = 9.8 m s^{- 2}

Solution:

$T = 2 π \frac{x _{0}}{g} = 2 π \frac{0.098}{9.8} = 2 π (0.1)$ .

Answer: $T \approx 0.63 s$ .

Solved Example 15

A block has period

3 s

on spring A and

4 s

on spring B. Find its period when the springs are joined (a) in series, (b) in parallel.

Solution:

$k \propto 1/ T^{2}$ . (a) Series: $T^{2} = T_{1}^{2} + T_{2}^{2} = 9 + 16$ , so $T = 5 s$ . (b) Parallel: $\frac{1}{T ^{2}} = \frac{1}{9} + \frac{1}{16} = \frac{25}{144}$ , so $T = 2.4 s$ .

Answer: $5 s$ in series, $2.4 s$ in parallel.

Solved Example 16

A mass

m

on a spring of constant

k

has period

T

. The spring is cut into two equal halves, the halves are joined in parallel and the same mass is attached. The new period is
(A)

2 T

(B)

T

(C)

T /2

(D)

T / 2

Solution:

Each half has constant $2 k$ ; in parallel, $k_{e q} = 4 k$ . $T^{'} = 2 π m /4 k = T /2$ .

Answer: (C).

Solved Example 17

A block on a vertical spring has period

T

. It is moved to a spring twice as stiff, and the whole set-up is placed in a lift accelerating upwards at

g /2

. The new period is
(A)

T / 2

(B)

T / 3

(C)

T

(D)

2 T

Solution:

The lift only changes the effective $g$ , which shifts the mean position ( $x_{0} = m g_{eff} / k$ ) but not the period. Only $m$ and $k$ matter: $T^{'} = 2 π \frac{m}{2 k} = \frac{T}{2} \approx 0.71 T$ .

Answer: (A). Option (B) wrongly treats the spring like a pendulum with $g_{eff} = 3 g /2$ .

Practice Questions

A $2 kg$ block oscillates on a spring of constant $200 N m^{- 1}$ . Find the period.Answer: $2 π 0.01 \approx 0.63 s$ .
A spring is cut into three equal parts and a mass $m$ is hung from one part. How does the period compare with the full spring?Answer: $k \to 3 k$ , so $T \to T / 3$ .
Springs $k$ and $2 k$ are joined in series and carry a mass $m$ . Find the period.Answer: $k_{e q} = 2 k /3$ , so $T = 2 π 3 m /2 k$ .
A $1 kg$ block stretches a spring by $4 cm$ . Find the frequency of vertical oscillation ( $g = 10 m s^{- 2}$ ).Answer: $f = \frac{1}{2 π} g / x_{0} = \frac{250}{2 π} \approx 2.5 Hz$ .
Blocks of $2 kg$ and $3 kg$ are joined by a spring of constant $600 N m^{- 1}$ on a smooth floor. Find the period.Answer: $μ = 1.2 kg$ ; $T = 2 π 1.2/600 \approx 0.28 s$ .
In Solved Example 12, take $M = 3 kg$ , $m = 1 kg$ , $k = 100 N m^{- 1}$ , $μ = 0.4$ , $g = 10 m s^{- 2}$ . Find the maximum amplitude without slipping.Answer: $A = μ (M + m) g / k = 0.16 m$ .
A $10 g$ bullet moving at $400 m s^{- 1}$ embeds in a $0.99 kg$ block at rest on a spring of constant $100 N m^{- 1}$ on a smooth floor. Find the amplitude.Answer: $v = 4 m s^{- 1}$ , $ω = 10 s^{- 1}$ , $A = 0.4 m$ .

Common Mistakes to Avoid

Watch out

Measuring displacement from the natural length in a vertical spring. SHM is about the stretched mean position $x_{0} = m g / k$ .
Thinking gravity changes the period of a spring-block. It only shifts the mean position; $T = 2 π m / k$ even on the Moon.
Adding spring constants in series. In series the reciprocals add, so $k_{e q}$ is smaller than either spring.
Halving $k$ when a spring is cut in half. Each half is stiffer: $2 k$ .
After a collision or added mass, using the old mean position to find the amplitude. Find the new mean position first.
Using energy conservation across an inelastic collision. Use momentum for the collision, energy only afterwards.
Assuming the block and spring always move by the same amount when a pulley is involved. Check the string constraint ( $x /2$ or $2 x$ ).
Using $m_{1} + m_{2}$ instead of the reduced mass $μ$ for two blocks joined by a spring.

Frequently Asked Questions

What is the time period of a spring-mass system?

The time period of a spring-mass system is $T = 2 π m / k$ , where $m$ is the mass and $k$ the spring constant. It does not depend on the amplitude or on gravity, so a horizontal, vertical or inclined spring with the same mass and spring has the same period.

Does gravity affect the time period of a vertical spring?

No. Gravity stretches the spring to a new mean position where the spring force balances the weight, but measured from that position the restoring force is still minus k x. The period stays two pi root m by k. Only the mean position moves down by m g over k.

What happens to the time period when a spring is cut in half?

Each half has twice the spring constant, because $k ℓ$ is constant for a given spring. With the same mass the period becomes $T / 2$ . If the two halves are then joined in parallel, the constant becomes $4 k$ and the period halves.

How do you find the equivalent spring constant in series and parallel?

In series the springs carry the same force, so the reciprocals add: one over k equivalent equals one over k1 plus one over k2. In parallel they have the same extension, so the constants add: k equivalent equals k1 plus k2. A block between two walls is a parallel case.

What is reduced mass in a two-block spring system?

When two free blocks are joined by a spring, both oscillate with the same period $T = 2 π μ / k$ , where the reduced mass is $μ = m_{1} m_{2} / (m_{1} + m_{2})$ . The centre of mass stays at rest, and the amplitudes divide inversely as the masses.

Does the time period of a spring-mass system depend on amplitude?

No. For an ideal spring obeying Hooke's law the period depends only on mass and spring constant. A larger amplitude means larger speeds, so the block covers the longer path in the same time. The energy, one half k A squared, does grow with amplitude.

Which spring-mass SHM questions are common in JEE Main and JEE Advanced?

JEE favours springs with pulleys, collisions followed by SHM, blocks that must not lift or separate, friction between stacked blocks, two-block systems with reduced mass, and energy-method problems with rotating pulleys. The beta squared rule and the new-mean-position routine solve most of them quickly.

What spring-mass questions come in NEET?

NEET asks direct questions: the period two pi root m by k, the effect of doubling mass or spring constant, series and parallel springs, cutting a spring into pieces, and the period from the static stretch of a vertical spring. Remember T squared adds in series and one over T squared adds in parallel.

Previous year questions on SHM in Spring-Mass System

10 questions from past papers, each with a step-by-step solution.

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