Miscellaneous Examples of SHM

Q: How do you add two simple harmonic motions of the same frequency?

Treat each amplitude as a vector at its phase angle and add them head to tail. The result is SHM of the same frequency with amplitude √(A₁² + A₂² + 2A₁A₂ cos δ), where δ is the phase difference. First convert both motions to the sine form.

Q: Is the sum of two SHMs always SHM?

Only if both have the same frequency and act along the same line. With different frequencies the sum is periodic (if the ratio of frequencies is rational) but not simple harmonic, because acceleration is no longer proportional to displacement. At right angles the particle moves on a line, ellipse or circle.

Q: What are Lissajous figures?

Lissajous figures are the paths traced when a particle performs two SHMs at right angles. For equal frequencies they are straight lines, ellipses or circles depending on the phase difference. For unequal frequencies in a whole-number ratio they form closed loops such as the figure of eight.

Q: What is the time period of liquid oscillating in a U-tube?

For a liquid column of total length L in a uniform U-tube, T = 2π√(L/2g). The extra column of height 2y supplies the restoring force. The period does not depend on the liquid's density or the tube's cross-section.

Q: How long would it take to fall through a tunnel through the Earth?

Inside a uniform Earth gravity is proportional to distance from the centre, so a body in a smooth tunnel performs SHM with period two pi root R over g, about 84.6 minutes. It reaches the other end in half a period, about 42 minutes, whatever the chord.

Q: Why does a floating body oscillate in SHM?

When pushed down by x, a floating body of cross-section A gains extra buoyancy Aσgx proportional to x and directed upwards. That is a restoring force of the SHM type, giving T = 2π√(d/g) with d the immersed depth at rest.

Q: Which miscellaneous SHM questions come in JEE Advanced?

JEE Advanced likes combinations of SHMs with the phasor method, perpendicular SHMs and their paths, the Earth tunnel along a chord, rolling bodies in bowls, gas-cushioned pistons with adiabatic changes, and checking whether a given function represents SHM. The restoring-force test solves all of them.

Q: Are these miscellaneous SHM topics in the NEET syllabus?

NEET mainly asks the resultant amplitude of two SHMs, whether an expression like sine squared is SHM, and simple systems such as the U-tube and floating body. Detailed Lissajous figures, rolling in bowls and gas pistons are more typical of JEE, but the formulas are quick to learn.

PhysicsOscillationsFor JEE aspirants

These miscellaneous examples of SHM go beyond springs and pendulums. Two SHMs of the same frequency along one line add like vectors into another SHM, $A^{2} = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos δ$ ; at right angles they trace straight lines, ellipses, circles and Lissajous figures. The same "restoring force $= - k x$ " test also gives the periods of a liquid in a U-tube, a floating body, a tunnel through the Earth, a ball in a bowl and a gas-cushioned piston. Miscellaneous SHM examples are a favourite source of JEE Advanced and JEE Main questions.

On this page1Same-line superposition2Is it SHM?3Perpendicular SHMs4U-tube5Floating body6Earth tunnel7Bowl8Gas piston9Mind map

Key Formulas - Quick Reference

★ Must learnSame line, same $ω$ : $x = A sin (ω t + ε)$ with $A = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos δ$
$tan ε = \frac{A _{2} sin δ}{A _{1} + A _{2} cos δ}$ (measured from the phase of $A_{1}$ )
$δ = 0$ : $A = A_{1} + A_{2}$ ; $δ = π$ : $A = ∣ A_{1} - A_{2} ∣$ ; $δ = π /2$ : $A = A_{1}^{2} + A_{2}^{2}$
Different frequencies: periodic (if $ω_{1} / ω_{2}$ is rational) but not SHM
★ Must learnPerpendicular, same $ω$ : $\frac{x ^{2}}{A _{1}^{2}} + \frac{y ^{2}}{A _{2}^{2}} - \frac{2 x y cos δ}{A _{1} A _{2}} = sin^{2} δ$
★ Must learnU-tube: $T = 2 π \frac{L}{2 g}$ ( $L$ = total length of the liquid column)
Floating body: $T = 2 π \frac{d}{g}$ ( $d$ = depth immersed at rest)
★ Must learnTunnel through a uniform Earth (any chord): $T = 2 π R / g \approx 84.6 min$
Ball in a bowl: $T = 2 π \frac{R - r}{g}$ (sliding), $2 π \frac{7 ( R - r )}{5 g}$ (rolling solid sphere)
Gas piston: $T = 2 π \frac{m L}{P _{0} A}$ (isothermal); replace $P_{0}$ by $γ P_{0}$ if adiabatic

1. Combining Two SHMs Along the Same Line

If two restoring forces, each proportional to displacement, act on a particle, the resulting motion is the combination (superposition) of two simple harmonic motions: the displacements simply add.

1.1 Same frequency: the vector (phasor) method

Let $x_{1} = A_{1} sin ω t$ and $x_{2} = A_{2} sin (ω t + δ)$ . Expanding and collecting the $sin ω t$ and $cos ω t$ terms:

x = (A_{1} + A_{2} cos δ) sin ω t + (A_{2} sin δ) cos ω t = A sin (ω t + ε)

A = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos δ, tan ε = \frac{A _{2} sin δ}{A _{1} + A _{2} cos δ}

The result is again SHM with the same frequency. The formulas are exactly those of vector addition: draw each amplitude as a vector at its phase angle and add head to tail. The method extends to three or more SHMs.

Figure 1: Treat each amplitude as a vector at its phase angle. For

A_{1} = 3

A_{2} = 4

δ = 9 0^{\circ}

: resultant

A = 5

ε = 5 3^{\circ}

Figure 2:

x_{1} = 3 sin ω t

(grey),

x_{2} = 4 cos ω t

(amber) and

x = x_{1} + x_{2}

(orange). The sum is again a sine curve with the same period, amplitude

5

: still SHM.

1.2 Rules before you add

Convert every term into the sine form, with $ω t$ carrying a positive sign ( $cos ω t = sin (ω t + π /2)$ , $- sin ω t = sin (ω t + π)$ ).
Make every amplitude positive (absorb a minus sign as $+ π$ in the phase).
Take $A_{1}$ as the amplitude of the SHM with the smaller phase; $δ$ is the phase difference.
Write the result as $x = A_{net} sin (phase of A_{1} + ε)$ .

Phase difference $δ$	Resultant amplitude	Remark
$0$ (in phase)	$A_{1} + A_{2}$	maximum
$π /2$	$A_{1}^{2} + A_{2}^{2}$	vectors at right angles
$2 π /3$ with $A_{1} = A_{2} = a$	$a$	equal SHMs $12 0^{\circ}$ apart
$π$ (opposite)	$∣ A_{1} - A_{2} ∣$	minimum; zero if equal

1.3 Different frequencies

If $x = A_{1} sin ω_{1} t + A_{2} sin ω_{2} t$ with $ω_{1} \neq = ω_{2}$ , the sum is not SHM: its acceleration is not proportional to $- x$ . It is periodic when $ω_{1} / ω_{2}$ is a ratio of whole numbers, with a period equal to the least common multiple of the two periods. (Close frequencies give beats, studied with waves.)

Figure 3:

x = sin ω t + sin 2 ω t

repeats every

T = 2 π / ω

, but the shape is not a sine curve: periodic, not SHM.

Exam Trick

"Is it SHM?" in ten seconds. Reduce the expression to a sum of sines and cosines. One frequency only (plus possibly a constant) means SHM; a constant only shifts the mean position.

Expression	Rewritten	SHM?
$sin ω t + cos ω t$	$2 sin (ω t + π /4)$	yes, amplitude $2$
$sin^{2} ω t$	$\frac{1}{2} - \frac{1}{2} cos 2 ω t$	yes, about $x = \frac{1}{2}$ , frequency $2 ω$
$sin ω t cos ω t$	$\frac{1}{2} sin 2 ω t$	yes, frequency $2 ω$
$sin^{3} ω t$	$\frac{3}{4} sin ω t - \frac{1}{4} sin 3 ω t$	no (two frequencies), periodic
$sin ω t + sin 2 ω t$	two frequencies	no, periodic
$e^{- ω t} sin ω t$	decaying	no, not periodic (damped)

Figure 4:

x = A sin^{2} ω t = \frac{A}{2} - \frac{A}{2} cos 2 ω t

: SHM about

x = A /2

with amplitude

A /2

, angular frequency

2 ω

and period

π / ω

(Solved Example 6). Squaring halved the period.

Key idea

Same line and same frequency: the sum is always SHM of that frequency. Any second frequency spoils it.

2. Two SHMs at Right Angles

Let $x = A_{1} sin ω t$ and $y = A_{2} sin (ω t + δ)$ , same frequency, perpendicular directions. Eliminate $t$ : $sin ω t = x / A_{1}$ and $cos ω t = 1 - x^{2} / A_{1}^{2}$ . Substituting in $y / A_{2} = sin ω t cos δ + cos ω t sin δ$ and squaring gives the path

\frac{x ^{2}}{A _{1}^{2}} + \frac{y ^{2}}{A _{2}^{2}} - \frac{2 x y cos δ}{A _{1} A _{2}} = sin^{2} δ

which is, in general, an ellipse.

Same line

Displacements add as numbers: $x = x_{1} + x_{2}$ . Same $ω$ gives SHM with $A = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos δ$ .

At right angles

Displacements add as vectors: the particle moves in a plane. Same $ω$ gives a line, an ellipse or a circle, not a 1-D SHM.

$δ$	Path	Equation
$0$	straight line through the origin, positive slope	$y = \frac{A _{2}}{A _{1}} x$
$π$	straight line, negative slope	$y = - \frac{A _{2}}{A _{1}} x$
$π /2$	ellipse along the axes	$\frac{x ^{2}}{A _{1}^{2}} + \frac{y ^{2}}{A _{2}^{2}} = 1$
$π /2$ and $A_{1} = A_{2} = A$	circle (uniform circular motion)	$x^{2} + y^{2} = A^{2}$
any other value	tilted ellipse	general equation above

Figure 5:

x = A_{1} sin ω t

y = A_{2} sin (ω t + δ)

. Lines at

δ = 0, π

; ellipses in between; arrows show the (clockwise) sense for

0 < δ < π

. With

A_{1} = A_{2}

and

δ = 9 0^{\circ}

the ellipse is a circle.

Exam Trick

Sense of rotation. For $x = A_{1} sin ω t$ , $y = A_{2} sin (ω t + δ)$ : if $0 < δ < π$ the point moves clockwise; if $π < δ < 2 π$ , anticlockwise. Check quickly by finding the direction of motion at $t = 0$ .

These paths are called Lissajous figures. With unequal frequencies the figures become more intricate:

Figure 6: Lissajous figures for unequal frequencies

ω_{x} : ω_{y}

. The path closes only when the ratio is a ratio of whole numbers; the motion is periodic but the path is not an ellipse.

Quick Recall: tap to check

Two perpendicular SHMs of equal amplitude and frequency differ in phase by

π /2

. What is the path?

A circle.

What is the path for

δ = 0

A straight line

y = (A_{2} / A_{1}) x

x = sin^{2} ω t

SHM?

Yes:

x = \frac{1}{2} - \frac{1}{2} cos 2 ω t

, SHM about

x = \frac{1}{2}

with angular frequency

2 ω

Figure 7: Two questions sort every combination. Only same frequency + same line gives SHM again, with

A = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos δ

3. SHM in Other Systems

Every system below is solved with the same test: displace by a small $x$ , find the restoring force, write it as $F = - k_{eff} x$ , identify the moving mass $m_{eff}$ , then $T = 2 π m_{eff} / k_{eff}$ .

3.1 Liquid in a U-tube

Liquid of density $ρ$ fills a length $L$ of a U-tube of cross-section $A$ . Mass $= L A ρ$ .
Let the level rise by $y$ on one side and fall by $y$ on the other. The unbalanced column of height $2 y$ weighs $2 y A ρ g$ and acts to restore the levels: $F = - (2 A ρ g) y$ .
$ω^{2} = \frac{2 A ρ g}{L A ρ} = \frac{2 g}{L}$ , so
$T = 2 π \frac{L}{2 g}$
It does not depend on the density or the cross-section.

Figure 8: The extra column of height

2 y

weighs

2 y A ρ g

and pushes the whole liquid (mass

L A ρ

) back:

ω^{2} = 2 g / L

T = 2 π L /2 g

3.2 Floating body

A body of mass $m$ and uniform cross-section $A$ floats upright in a liquid of density $σ$ with depth $d$ immersed: $m g = A d σ g$ .
Pushed down by a further $x$ , extra buoyancy $A x σ g$ acts upwards: $F = - (A σ g) x$ .
$ω^{2} = \frac{A σ g}{m} = \frac{g}{d}$ , so
$T = 2 π \frac{d}{g}$
For a cylinder of height $h$ and density $ρ$ , $d = h ρ / σ$ . (Viscous drag and the moving liquid are ignored.)

Figure 9: Pushing the cylinder down by

x

adds buoyancy

A x σ g

(restoring). With

d

the depth at rest,

T = 2 π d / g = 2 π h ρ / σ g

3.3 Tunnel through the Earth

Inside a uniform Earth, only the mass within radius $r$ attracts: $g (r) = g \frac{r}{R}$ .
In a smooth tunnel along a diameter, at distance $x$ from the centre $F = - \frac{m g}{R} x$ : SHM with $ω = g / R$ .
Along a chord, the gravity $\frac{m g r}{R}$ has a component $\frac{m g}{R} x$ along the chord, $x$ measured from its middle: the same $ω$ .
$T = 2 π \frac{R}{g} \approx 84.6 min$
A body dropped in takes $T /2 \approx 42 min$ to reach the other end, passing the centre at $v_{m a x} = g R \approx 7.9 km s^{- 1}$ .

Figure 10: Inside a uniform Earth, gravity grows linearly with distance from the centre, so along any straight smooth tunnel

F = - \frac{m g}{R} x

and

T = 2 π R / g \approx 84.6

min.

Same number, three places. $2 π R / g \approx 84.6$ min is also the period of a satellite skimming the Earth's surface and of an infinitely long pendulum. All three are governed by $g$ and $R$ alone.

3.4 Ball in a bowl

A small ball of radius $r$ at the bottom of a smooth spherical bowl of radius $R$ moves on a circle of radius $R - r$ about the bowl's centre: it is a pendulum of length $R - r$ , so $T = 2 π (R - r) / g$ . If the ball rolls without slipping, its rotational energy adds to the inertia.

Figure 11: A ball in a bowl swings like a pendulum of length

R - r

about the bowl's centre

O

. Sliding (smooth bowl):

T = 2 π (R - r) / g

; rolling solid sphere:

T = 2 π 7 (R - r) /5 g

JEE Advanced

Rolling ball. Kinetic energy $= \frac{1}{2} m v^{2} + \frac{1}{2} \cdot \frac{2}{5} m r^{2} (\frac{v}{r})^{2} = \frac{1}{2} \cdot \frac{7}{5} m v^{2}$ , so $m_{eff} = \frac{7}{5} m$ while the restoring force stays $- \frac{m g}{R - r} s$ . Hence $T = 2 π \frac{7 ( R - r )}{5 g}$ . For a ring or hollow cylinder replace $\frac{7}{5}$ by $1 + K^{2} / r^{2}$ .

3.5 Piston on a gas column

A piston of mass $m$ and area $A$ traps a gas column of length $L$ at equilibrium pressure $P_{0}$ (which already balances the atmosphere and the piston's weight).
Push the piston in by $x$ . Isothermally, $P_{0} A L = P (L - x) A$ , so the pressure rises by $Δ P \approx P_{0} \frac{x}{L}$ .
Restoring force $F = - \frac{P _{0} A}{L} x$ , so
$T = 2 π \frac{m L}{P _{0} A}$
For rapid (adiabatic) compression $Δ P = γ P_{0} x / L$ and $P_{0} \to γ P_{0}$ .

Figure 12: Pushing the piston down by

x

raises the gas pressure by

P_{0} x / L

(isothermal), a restoring force

P_{0} A x / L

T = 2 π m L / P_{0} A

. For a quick (adiabatic) squeeze use

γ P_{0}

System	$k_{eff}$	$m_{eff}$	Period
U-tube	$2 A ρ g$	$L A ρ$	$2 π L /2 g$
Floating body	$A σ g$	$m = A d σ$	$2 π d / g$
Earth tunnel	$m g / R$	$m$	$2 π R / g$
Ball in bowl (sliding)	$m g / (R - r)$	$m$	$2 π (R - r) / g$
Gas piston (isothermal)	$P_{0} A / L$	$m$	$2 π m L / P_{0} A$

Key idea

Density, cross-section and mass often cancel: U-tube

2 π L /2 g

, floating body

2 π d / g

, Earth tunnel

2 π R / g

Quick Recall: tap to check

Does the period of liquid in a U-tube depend on the liquid's density?

No:

T = 2 π L /2 g

depends only on the column length.

How long does a ball take to fall through a tunnel along any chord of the Earth?

Half a period, about

42 min

, whatever the chord.

A piston is pushed in quickly (adiabatic). What replaces

P_{0}

T

γ P_{0}

, so the period becomes shorter.

4. The Whole Chapter on One Page

Figure 13: The whole chapter on one screen. Every branch uses the same idea: find the restoring force or torque, write it as

- k x

- k θ

, and read off

ω

Key idea

One test solves every SHM problem: displace slightly, find the restoring force or torque, write it as

- k x

- k θ

, and

ω = k / m

k / I

5. Solved Examples

Solved Example 1

x_{1} = 3 sin ω t

and

x_{2} = 4 cos ω t

. Find (i) the amplitude of the resultant SHM, (ii) the equation of the resultant SHM.

Solution:

Write both in sine form: $x_{1} = 3 sin ω t$ , $x_{2} = 4 sin (ω t + π /2)$ , so $δ = π /2$ .

$A = 3^{2} + 4^{2} + 2 (3) (4) cos 9 0^{\circ} = 25 = 5$ . $tan ε = \frac{4 sin 9 0 ^{\circ}}{3 + 4 cos 9 0 ^{\circ}} = \frac{4}{3}$ , so $ε \approx 5 3^{\circ}$ .

Answer: $A = 5$ ; $x = 5 sin (ω t + 5 3^{\circ})$ (Figures 1 and 2).

Solved Example 2

x_{1} = 5 sin (ω t + 3 0^{\circ})

and

x_{2} = 10 cos ω t

. Find the amplitude of the resultant SHM.

Solution:

$x_{2} = 10 sin (ω t + 9 0^{\circ})$ , so the phase difference is $9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ .

$A = 5^{2} + 1 0^{2} + 2 (5) (10) cos 6 0^{\circ} = 25 + 100 + 50 = 175$ .

Answer: $A = 57 \approx 13.2$ .

Solved Example 3

A particle is subjected to two SHMs

x_{1} = A_{1} sin ω t

and

x_{2} = A_{2} sin (ω t + π /3)

. Find (a) the displacement at

t = 0

, (b) the maximum speed, (c) the maximum acceleration.

Solution:

(a) At $t = 0$ : $x_{1} = 0$ , $x_{2} = A_{2} sin \frac{π}{3} = \frac{3}{2} A_{2}$ . So $x = \frac{3}{2} A_{2}$ .

The resultant is SHM of the same $ω$ with $A = A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos (π /3) = A_{1}^{2} + A_{2}^{2} + A_{1} A_{2}$ .

(b) $v_{m a x} = A ω = ω A_{1}^{2} + A_{2}^{2} + A_{1} A_{2}$ . (c) $a_{m a x} = A ω^{2} = ω^{2} A_{1}^{2} + A_{2}^{2} + A_{1} A_{2}$ .

Solved Example 4

Three SHMs of equal amplitude

a

and equal frequency act along the same line with phases

0

2 π /3

and

4 π /3

. The resultant amplitude is
(A)

3 a

(B)

a 3

(C)

a

(D)

0

Solution:

The three amplitude vectors are $12 0^{\circ}$ apart and equal: head to tail they form a closed equilateral triangle.

Answer: (D). The particle stays at rest.

Solved Example 5

A particle moves so that

x = 3 sin ω t

and

y = 4 cos ω t

. Find the path and the sense of motion.

Solution:

$\frac{x}{3} = sin ω t$ , $\frac{y}{4} = cos ω t$ , so $\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1$ : an ellipse with semi-axes $3$ and $4$ .

At $t = 0$ the particle is at $(0, 4)$ and $x$ is increasing, so it moves right from the top: clockwise. (Here $δ = π /2$ , inside $0$ to $π$ .)

Solved Example 6

Show that

x = A sin^{2} ω t

is SHM and find its mean position, amplitude and period.

Solution:

$x = \frac{A}{2} - \frac{A}{2} cos 2 ω t$ . So $x - \frac{A}{2} = - \frac{A}{2} cos 2 ω t$ and $\frac{d ^{2} x}{d t ^{2}} = - (2 ω)^{2} (x - \frac{A}{2})$ .

Answer: SHM about $x = A /2$ , amplitude $A /2$ , angular frequency $2 ω$ , period $\frac{π}{ω}$ .

Solved Example 7

A U-tube contains a liquid column of total length

40 cm

. The liquid is disturbed. Find the period of oscillation (

g = 10 m s^{- 2}

Solution:

$T = 2 π \frac{L}{2 g} = 2 π \frac{0.40}{20} = 2 π (0.141)$ .

Answer: $T \approx 0.89 s$ .

Solved Example 8

A wooden cylinder of density

600 kg m^{- 3}

and height

20 cm

floats upright in water. It is pushed down slightly and released. Find the period (

g = 9.8 m s^{- 2}

Solution:

Depth immersed $d = h \frac{ρ}{σ} = 20 \times \frac{600}{1000} = 12 cm$ .

$T = 2 π \frac{d}{g} = 2 π \frac{0.12}{9.8}$ . Answer: $T \approx 0.70 s$ .

Solved Example 9

A straight smooth tunnel is dug along a diameter of the Earth (radius

6.4 \times 1 0^{6} m

g = 9.8 m s^{- 2}

). A ball is dropped into it. Find (a) the time to reach the other end, (b) the speed at the centre.

Solution:

$ω = g / R$ ; $T = 2 π \frac{6.4 \times 1 0 ^{6}}{9.8} \approx 5.08 \times 1 0^{3} s \approx 84.6 min$ .

(a) From one end to the other is half a period: about $42 min$ . (b) $v_{m a x} = A ω = R g / R = g R \approx 7.9 km s^{- 1}$ .

Solved Example 10

A piston of mass

1 kg

and area

10 cm^{2}

traps a gas column

50 cm

long at equilibrium pressure

1.0 \times 1 0^{5} Pa

in a smooth cylinder. Find the period of small oscillations, assuming isothermal changes.

Solution:

$k_{eff} = \frac{P _{0} A}{L} = \frac{( 1.0 \times 1 0 ^{5} ) ( 1.0 \times 1 0 ^{- 3} )}{0.50} = 200 N m^{- 1}$ .

$T = 2 π \frac{m}{k _{eff}} = 2 π \frac{1}{200}$ . Answer: $T \approx 0.44 s$ .

Solved Example 11

A solid ball of radius

0.10 m

rolls without slipping inside a spherical bowl of radius

0.80 m

. Find the period of small oscillations (

g = 9.8 m s^{- 2}

Solution:

$T = 2 π \frac{7 ( R - r )}{5 g} = 2 π \frac{7 ( 0.70 )}{5 ( 9.8 )} = 2 π 0.10$ .

Answer: $T \approx 1.99 s$ .

Solved Example 12

Liquid in a U-tube oscillates with period

T

. It is replaced by a liquid of half the density, keeping the same total length of the liquid column. The new period is
(A)

T / 2

(B)

T

(C)

2 T

(D)

2 T

Solution:

Restoring force $= 2 A ρ g y$ and moving mass $= L A ρ$ : the density cancels, $T = 2 π \frac{L}{2 g}$ . Same $L$ means the same period.

Answer: (B). Halving the density halves both the restoring force and the mass.

Practice Questions

Find the amplitude of $x = 6 sin ω t + 8 sin (ω t + π)$ .Answer: $∣6 - 8∣ = 2$ .
Find the amplitude and phase of $x = sin ω t + 3 cos ω t$ .Answer: $2$ ; $x = 2 sin (ω t + π /3)$ .
Is $x = sin ω t cos 2 ω t$ SHM?Answer: No: $\frac{1}{2} (sin 3 ω t - sin ω t)$ has two frequencies.
$x = 2 sin ω t$ , $y = 2 sin (ω t + π)$ . Describe the path.Answer: The straight line $y = - x$ between $(- 2, 2)$ and $(2, - 2)$ .
A U-tube has mercury of total length $L$ . If the mercury is replaced by water of the same column length, how does the period change?Answer: It stays the same: $T = 2 π L /2 g$ does not depend on density.
A cube floats with $4 cm$ immersed. Find its period of vertical oscillation ( $g = π^{2} m s^{- 2}$ ).Answer: $2 π 0.04/ π^{2} = 0.4 s$ .
On a planet of half the Earth's radius and the same density, how does the tunnel period compare?Answer: The same: $R / g$ is unchanged because $g \propto ρR$ .

Common Mistakes to Avoid

Watch out

Adding amplitudes directly. Amplitudes add like vectors: $A_{1} + A_{2}$ only when $δ = 0$ .
Using $δ$ without converting $cos$ terms to $sin$ first. $cos ω t = sin (ω t + π /2)$ .
Ignoring a minus sign: $- sin ω t$ is $sin (ω t + π)$ , which changes $δ$ by $π$ .
Calling the sum of two different frequencies SHM. It is periodic at best, never simple harmonic.
Taking the U-tube column length as one arm. $L$ is the total length of liquid.
Using the whole height of a floating body in $T = 2 π d / g$ . Use the immersed depth $d$ .
Forgetting rotational kinetic energy for a ball rolling in a bowl (the factor $7/5$ ).
Measuring $x$ from the Earth's centre in a chord tunnel. Along the chord, $x$ is measured from the chord's midpoint.

Frequently Asked Questions

How do you add two simple harmonic motions of the same frequency?

Treat each amplitude as a vector at its phase angle and add them head to tail. The result is SHM of the same frequency with amplitude $A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} cos δ$ , where $δ$ is the phase difference. First convert both motions to the sine form.

Is the sum of two SHMs always SHM?

Only if both have the same frequency and act along the same line. With different frequencies the sum is periodic (if the ratio of frequencies is rational) but not simple harmonic, because acceleration is no longer proportional to displacement. At right angles the particle moves on a line, ellipse or circle.

What are Lissajous figures?

Lissajous figures are the paths traced when a particle performs two SHMs at right angles. For equal frequencies they are straight lines, ellipses or circles depending on the phase difference. For unequal frequencies in a whole-number ratio they form closed loops such as the figure of eight.

What is the time period of liquid oscillating in a U-tube?

For a liquid column of total length $L$ in a uniform U-tube, $T = 2 π L /2 g$ . The extra column of height $2 y$ supplies the restoring force. The period does not depend on the liquid's density or the tube's cross-section.

How long would it take to fall through a tunnel through the Earth?

Inside a uniform Earth gravity is proportional to distance from the centre, so a body in a smooth tunnel performs SHM with period two pi root R over g, about 84.6 minutes. It reaches the other end in half a period, about 42 minutes, whatever the chord.

Why does a floating body oscillate in SHM?

When pushed down by $x$ , a floating body of cross-section $A$ gains extra buoyancy $A σ g x$ proportional to $x$ and directed upwards. That is a restoring force of the SHM type, giving $T = 2 π d / g$ with $d$ the immersed depth at rest.

Which miscellaneous SHM questions come in JEE Advanced?

JEE Advanced likes combinations of SHMs with the phasor method, perpendicular SHMs and their paths, the Earth tunnel along a chord, rolling bodies in bowls, gas-cushioned pistons with adiabatic changes, and checking whether a given function represents SHM. The restoring-force test solves all of them.

Are these miscellaneous SHM topics in the NEET syllabus?

NEET mainly asks the resultant amplitude of two SHMs, whether an expression like sine squared is SHM, and simple systems such as the U-tube and floating body. Detailed Lissajous figures, rolling in bowls and gas pistons are more typical of JEE, but the formulas are quick to learn.

Previous year questions on Miscellaneous Examples of SHM

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

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