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
$F=-kx$
Hence, by second law equation for $m_1$ and $m_2$ are
$m_1{a}_1=-kx$
$\Rightarrow m_1\frac{d^2{x}_1}{d{t}^2}=-kx$
and $m_2{a}_2=-kx$
$\Rightarrow m_2\frac{d^2{x}_2}{d{t}^2}=-kx$
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
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{1}{k}\left(\frac{m_1{m}_2}{m_1+m_2}\right)}$