SHM in Spring-Mass System

SIMPLE HARMONIC MOTION IN SPRING-MASS SYSTEM

Let us find out the time period of a spring-mass system oscillating on a smooth horizontal surface as shown in the figure.

At the equilibrium position the spring is relaxed. When the block is displaced through a distance x towards right, it experiences a net restoring force F = –kx towards left.

The negative sign shows that the restoring force is always opposite to the displacement. That is, when x is positive, F is negative, the force is directed to the left. When x is negative, F is positive, the force is directed to the right. Thus, the force always tends to restore the block to its equilibrium position x = 0.

F = -kx

Applying Newton's Second Law,

$F=m\frac{d^{2}x}{d{t}^2}=-kx$

or $\frac{d^{2}x}{d{t}^2}+\left(\frac{k}{m}\right)x=0$

Comparing the above equation with, we get

${\omega }^2=\frac{k}{m}$

or $T=2\pi \sqrt{\frac{m}{k}}$

Exmaple 1: Find the period of oscillation of a vertical spring-mass system.

Solution: Let $x_0$be the deformation in the spring in equilibrium.

Then $k{x}_0=mg$

When the block is further displaced by x, the net restoring force is given by

$F=-\left[k\left(x+x_0\right)-mg\right]$

or $F=-kx\left(becausek{x}_0=mg\right)$

Using second law of motion

$m\frac{d^{2}x}{d{t}^2}=-kx$

or $\frac{d^{2}x}{d{t}^2}+\left(\frac{k}{m}\right)x=0$

Thus, ${\omega }^2=\frac{k}{m}orT=2\pi \sqrt{\frac{m}{k}}$

Series and Parallel Combinations of Springs

When two springs are joined in series, the equivalent stiffness of the combination may obtained as

$\frac{1}{k}=\frac{1}{k_1}+\frac{1}{k_2}$

When two springs are joined in parallel, the equivalent stiffness of the combination is given by

$k=k_1+k_2$

Fig. (a) Series combination of spring (b) Parallel combination of spring

Exmaple 2: A spring of stiffness constant k and natural length $\ell$ is cut into two parts of length $\frac{3\ell }{4}$ and $\frac{\ell }{4}$ respectively, and an arrangement is made as shown in the figure. If the mass is slightly displaced, find the time period of oscillation.

Solution: The stiffness of a spring is inversely proportional to its length. Therefore the stiffness of each part is $k_1=\frac{4}{3}k$ and $k_2=4k$

Time period, $T=2\pi \sqrt{\frac{m}{k_1+k_2}}$

or $T=2\pi \sqrt{\frac{3m}{16k}}=\frac{\pi }{2}\sqrt{\frac{3m}{k}}$

Exmaple 3: Two masses $m_1$and $m_2$ are connected by a spring of force constant k and are placed on a frictionless horizontal surface. Show that if the masses are displaced slightly in opposite directions and released, the system will execute simple harmonic motion. Calculate the frequency of oscillation.

Solution: Let masses $m_1$and $m_2$ be displaced by $x_1$and $x_2$respectively from their equilibrium position in opposite direction so that the total extension in the spring will be $x=x_1+x_2$. Due to this stretch a restoring force kx will act on each mass and so equation of mass $m_1$ will be

$m_1\frac{d^2{x}_1}{d{t}^2}=-kx,$ i.e. $\frac{d^2{x}_1}{d{t}^2}=-\frac{k}{m_1}x$ … (i)

while that for m₂ will be

$m_2\frac{d^2{x}_2}{d{t}^2}=-kx,$i. e. $\frac{d^2{x}_2}{d{t}^2}=-\frac{k}{m_2}x$ … (ii)

But as x = x₁ + x₂ i.e. $\frac{d^2{x}_2}{d{t}^2}=\frac{d^2{x}_1}{d{t}^2}+\frac{d^2{x}_2}{d{t}^2}$ … (iii)

So substituting equation (i) and (ii) in (iii),

$\frac{d^2{x}_2}{d{t}^2}=-\left(\frac{1}{m_1}+\frac{1}{m_2}\right)kx$

or $\frac{d^2{x}_2}{d{t}^2}=-\frac{k}{m}xwhere\frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2}$

or $\frac{d^2{x}_2}{d{t}^2}=-{\omega }^{2}xwhere{\omega }^2=\frac{k}{m}$

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}=\frac{1}{2\pi }\sqrt{\frac{k\left(m_1+m_2\right)}{m_1{m}_2}}$

SUPERPOSITION OF TWO SHM'S

(a) In same direction and of same frequency.

\begin{align} {{x}_{1}}={{A}_{1}}\sin \omega t \\ {{x}_{2}}={{A}_{2}}\sin \left( \omega t+\phi \right),\text{  then resultant displacement} \\ \end{align}

$x_1+x_2=A_1\sin \omega t+A_2\sin \left(\omega t+\theta \right)=A\sin \left(\omega t+\varphi \right)$

If $\theta =0$, both SHM's are in phase and $A=A_1+A_2$

If $\theta =\pi$, both SHM's are out of phase and $A=\left|A_1-A_2\right|$

The resultant amplitude due to superposition of two or more than two SHM's of this case can also be found by phasor diagram also.

(b) In same direction but are of different frequencies. (special case, if $A_1=A_2$)

\begin{align}  {{x}_{1}}={{A}_{1}}\sin {{\omega }_{1}}t \\  {{x}_{2}}={{A}_{2}}\sin {{\omega }_{2}}t \\ \end{align}

then resultant displacement $x=x_1+x_2=A_1\sin {\omega }_{1}t-A_2\sin {\omega }_{2}t$

(c) In two perpendicular directions

\begin{align}  x=A\sin \omega t \\  y=B\sin \left( \omega t+\phi \right) \\\end{align}

Case(i) if $\theta =0or\pi theny\pm \left(B/A\right)x.$ So path will be straight line & resultant displacement will be $r={\left(x^2+y^2\right)}^{1/2}={\left(A^2+B^2\right)}^{1/2}\sin \omega t$

Case (ii) if $\theta =\frac{\pi }{2}then.$

$x=A\sin \omega t$

$y=B\sin \left(\omega t+\pi 2\right)=B\cos \omega t$

so, resultant will be $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$. i.e. equation of an ellipse and if A = B, then superposition will be an equation of circle.