Question

The bob of a simple pendulum has a mass m and it is executing simple harmonic motion of amplitude A and period T. it collides with a body of mass $m_0$ placed at the equilibrium position which sticks to the bob. The time period of the oscillation of the combined masses will be

• A. $T$
• B. $T\frac{m+m_0}{m-m_0}$
• C. $T\frac{m}{m-m_0}$
• D. $T\frac{m+m_0}{m}$

Answer 1

The correct option is D $T\frac{m+m_0}{m}$

Let $A_0,T_0$ and ${\omega }_0$ respectively be the amplitude, time period and angular frequency of the combined masses. Let υ be the velocity of the bob when it reaches the equilibrium position and $V_0$ The velocity of the combined masses just after collision. Then, from the principle of conservation of momentm, we have
$mv=(m+m_0)V_0$ (1)
Now v=r $\omega =r\times \frac{2\pi }{T}$ and $V_0$ $r{\omega }_0=r\times \frac{2\pi }{T_0}$ where r is the length of the pendulum. Using these in (1), we get
$m\times \left(\frac{2\pi r}{T}\right)=(m+m_0)\times \left(\frac{2\pi r}{T_0}\right)$
or $T_0=T\frac{m+m_0}{m}$ which is choice (d).