Question

Two masses $m_1$ and $m_2$ connected by a spring of spring constant $k$ rest on a frictionless surface. If the masses are pulled apart and let go, the time period of oscillation is

• A. $T=2\pi \sqrt{\frac{1}{k}\left(\frac{m_{1}m_{2}}{m_{1}+m_{2}}\right)}$
• B. $T=2\pi \sqrt{k\left(\frac{m_{1}+m_{2}}{m_{1}+m_{2}}\right)}$
• C. $T=2\pi \sqrt{\frac{m_{1}}{k}}$
• D. $T=2\pi \sqrt{\frac{m_{2}}{k}}$

Answer 1

Answer: (A)

Let displacements of masses $m_{1}$ and $m_{2}$ are $x_{1}$ and $x_{2}$, respectively.

Total elongation of spring is $x=x_{1}+x_{2}$.
So, when spring snaps back, pull on each of mass is
$F=-k x$
Hence, by second law equation for $m_{1}$ and $m_{2}$ are
$m_{1} a_{1}=-k x $
$\Rightarrow m_{1} \frac{d^{2} x_{1}}{d t^{2}}=-k x$
and $m_{2} a_{2}=-k x $
$\Rightarrow m_{2} \frac{d^{2} x_{2}}{d t^{2}}=-k x$
Now, from $x=x_{1}+x_{2}$, we have
$\frac{d^{2} x}{d t^{2}}=\frac{d^{2} x_{1}}{d t^{2}}+\frac{d^{2} x_{2}}{d t^{2}}$
$\Rightarrow \frac{d^{2} x}{d t^{2}}=\frac{-k}{m_{1}} x+\frac{-k}{m_{2}} x$
$\Rightarrow \frac{d^{2} x}{d t^{2}}=-k\left(\frac{1}{m_{1}}+\frac{1}{m_{2}}\right) x$
So, $\omega^{2}=k\left(\frac{1}{m_{1}}+\frac{1}{m_{2}}\right)=k\left(\frac{m_{1}+m_{2}}{m_{1} m_{2}}\right)$
Hence, time period of oscillation is
$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{1}{k}\left(\frac{m_{1} m_{2}}{m_{1}+m_{2}}\right)}$