Question

The pendulum of a grandfather’s clock takes 1s to oscillate from one end to another a distance of 10 cm. Considering it a simple pendulum, find its maximum velocity
(A) 4 cm/s
(B) 8 cm/s
(C) 12 cm/s
(D) $ 15.7cm/s $

Answer 1

Hint The amplitude of a simple pendulum oscillation is the maximum distance travelled by the oscillator (the bob) from the equilibrium position. The equilibrium position is at the centre of the swing. The period is the total time taken to complete a full oscillation, i.e. the to and from movement of the bob.
Formula used: In this solution we will be using the following formulae;
 $ \Rightarrow y = A\sin \omega t $ where $ y $ is the distance covered by the oscillator from equilibrium position, $ A $ is the amplitude, $ \omega $ is the angular speed, and $ t $ is the time taken.
 $ \Rightarrow {v_{\max }} = A\omega $ , where $ {v_{\max }} $ is the maximum velocity of the oscillator.
 $ \Rightarrow \omega = \dfrac{{2\pi }}{T} $ where $ T $ is the period of oscillation.

Complete step by step answer
The motion of the grandfather’s clock is identical to a pendulum bob oscillatory motion. When neglecting air resistance, it can be described as
 $ \Rightarrow y = A\sin \omega t $ where $ y $ is the distance covered from equilibrium position after a time $ t $ by the oscillator, $ A $ is the amplitude, $ \omega $ is the angular speed.
The velocity of oscillation is given by differentiating the distance with respect to time, i.e.
 $ \Rightarrow \dfrac{{dy}}{{dt}} = A\omega \cos \omega t $
Now, just like amplitude is the maximum displacement from the equilibrium position, $ A\omega $ is the maximum velocity every attained during oscillation. Hence
 $ \Rightarrow {v_{\max }} = A\omega $
Now, the distance covered from one end to another is given as 10 cm but the amplitude is the distance from the centre (since that is the equilibrium position), Hence,
 $ \Rightarrow A = \dfrac{{10}}{2} = 5cm $
To calculate the angular speed $ \omega $ , we have that
 $ \Rightarrow \omega = \dfrac{{2\pi }}{T} $ where $ T $ is the period of oscillation.
Note that the time taken to go from one end to the other is given as 1 second, but the period is the time taken to go to and fro. Hence
 $ \Rightarrow T = 2t = 2\left( 1 \right) = 2s $
Hence, inserting into $ \omega = \dfrac{{2\pi }}{T} $ , we have
 $ \Rightarrow \omega = \dfrac{{2\pi }}{2} $
 $ \Rightarrow \omega = \pi {s^{ - 1}} $
Thus,
 $ \Rightarrow {v_{\max }} = A\omega = 5 \times \pi $
 $ \Rightarrow {v_{\max }} = 15.7cm/s $
Hence the correct option is D.

Note
In reality, no simple pendulum repeats itself the exact same way for each cycle. This is because of air resistance on the oscillator, and possibly the friction at the fixed point where the cord is tied. Hence, in such case, our deduction that $ T = 2t $ do not hold because even after a half cycle (say to), some energy has been dissipated by this resistant forces, and thus at the next half cycle (say fro), the oscillator has a lower speed and would hence take more time to complete that half cycle than it did to complete the first half cycle.