Change of Time period of a pendulum with additional mass

[Hijaz Aslam]

Homework Statement

Q. [/B]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 $m_o$ 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. $T \sqrt{\frac{m+m_o}{m-m_o}}$ c. $T \sqrt{\frac{m}{m-m_o}}$ d. $T \sqrt{\frac{m+m_o}{m}}$

Homework Equations

Time period of a simple pendulum = $2\pi \sqrt{\frac{l}{g}}$

The Attempt at a Solution

According to the time period equation, the 'time period ($T$)', 'frequency($\nu$)' and 'angular frequency($\omega$)' shall not change with the mass of the bob.

I think even when an external mass is added to the bob during motion the amplitude and other related factors (maximum velocity, energy etc) shall change except the Time Period.

But my text gives the answer as option (d).
It gives the solution as follows:

Let the velocity of the bob at the mean position be $v$ and the velocity of the combined mass $(m+m_o)$ be $v_o$. Then according to the conservation of energy : $$\frac{1}{2}mv^2=\frac{1}{2}(m+m_o)v_o^2$$

Now $$v=r\omega=r\frac{2\pi}{T}$$ and $$v_o=r\omega _o=r\frac{2\pi}{T_o}$$
Substituting the above two results in the former equation and simplifying the final answer should be : $$T \sqrt{\frac{m+m_o}{m}}$$I am simply baffled by how the text made these two statements $$r\omega=r\frac{2\pi}{T}$$ and $$r\omega _o=r\frac{2\pi}{T_o}$$. Here $\omega$ and $\omega _o$ are the angular velocities and not the 'angular frequency' of the whole pendulum isn't it? Is it a 'schoolboy error' done by my textbook or am I missing any concepts? (I took time to post this question because, the text am referring to, is well reputed for its error free methods.)

 

[DuckAmuck]

You text is wrong. Frequency will not change with mass.

 

[TSny]

I think even when an external mass is added to the bob during motion the amplitude and other related factors (maximum velocity, energy etc) shall change except the Time Period.

You are right, the time period will not change, as confirmed by DuckAmuk.

But my text gives the answer as option (d).
It gives the solution as follows:

Let the velocity of the bob at the mean position be $v$ and the velocity of the combined mass $(m+m_o)$ be $v_o$. Then according to the conservation of energy : $$\frac{1}{2}mv^2=\frac{1}{2}(m+m_o)v_o^2$$

The masses stick together. So, the collision is inelastic. The kinetic energy is not conserved.

Now $$v=r\omega=r\frac{2\pi}{T}$$ and $$v_o=r\omega _o=r\frac{2\pi}{T_o}$$
Substituting the above two results in the former equation and simplifying the final answer should be : $$T \sqrt{\frac{m+m_o}{m}}$$

I am simply baffled by how the text made these two statements $$r\omega=r\frac{2\pi}{T}$$ and $$r\omega _o=r\frac{2\pi}{T_o}$$. Here $\omega$ and $\omega _o$ are the angular velocities and not the 'angular frequency' of the whole pendulum isn't it?

Yes, that's exactly right. Angular velocity of the pendulum and angular frequency of the simple harmonic motion are two entirely different quantities. It's disconcerting that a textbook would make these errors. Do you mind telling us which textbook you are using?