Question

What would be the period of the free oscillations of the system shown here if mass $M_1$ is pulled down a little? Force constant of the spring is k, the mass of fixed pulley is negligible and movable pulley is smooth

• A. $T=2\pi \sqrt{\frac{M_1+M_2}{k}}$
• B. $T=2\pi \sqrt{\frac{M_1+4M_2}{k}}$
• C. $T=2\pi \sqrt{\frac{M_2+4M_1}{k}}$
• D. $T=2\pi \sqrt{\frac{M_2+3M_1}{k}}$

Answer 1

The correct option is C $T=2\pi \sqrt{\frac{M_2+4M_1}{k}}$

Let the system spring extended by $x$ at equilibrium position.

In this position tension of mass $m_1$ is $T^{\prime }$ and mass $m_2$ (pulley) be $2T^{\prime }$.

So, we have force equation as,

for $m_1$ $T=m_{1}g$

for $m_2$ $k{x}_0=2T-m_{2}g$

or, $k{x}_0=2m_{1}g-m_{2}g$ $---\left(i\right)$

Now consider displacement of mass $m_2$ displaced by $x$. The corresponding displacement for mass of block $m_1$ is $\left(2x\right)$. And in this position tension on $m_1$ be $T^{\prime \prime }$ and on $m_2$ be $2T^{\prime \prime }$.

So, for $m_2$: $m_2\frac{d^{2}x}{d{t}^2}=2T^{\prime \prime }-m_{2}g-kx-k{x}_0$ $---\left(ii\right)$

for $m_1$ : $m_1\frac{d^{2}2x}{d{t}^2}=m_{1}g-T^{\prime \prime }$

$\Rightarrow 2m_1\frac{d^{2}x}{d{t}^2}=m_{1}g-T^{\prime \prime }$ $---\left(iii\right)$

Multiplying eqn.$\left(iii\right)$ by $2$ and adding it with eqn.$\left(ii\right)$ we get,

$(m_2+4m_1)\frac{d^{2}x}{d{t}^2}=2m_{1}g-m_{2}g-kx-k{x}_0$

Putting the value of $k{x}_0$, we get,

$(m_2+4m_1)\frac{d^{2}x}{d{t}^2}=2m_{1}g-m_{2}g-kx-2m_{1}g+m_{2}g$

$\Rightarrow \frac{d^{2}x}{d{t}^2}+\left(\frac{k}{m_2+4m_1}\right)x=0$

So, angular frequency, $\omega =\sqrt{\frac{k}{m_2+4m_1}}$

Time period $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m_2+4m_1}{k}}$