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Question: lf a simple pendulum oscillates with an amplitude of \[50\,mm\] and time period of \[2{\text{ }}sec\...

lf a simple pendulum oscillates with an amplitude of 50 mm50\,mm and time period of 2 sec2{\text{ }}sec then its maximum velocity is:
A. 0.6 ms−10.6\,m{s^{ - 1}}
B. 0.16 ms−10.16\,m{s^{ - 1}}
C. 0.8 ms−10.8\,m{s^{ - 1}}
D. 0.32 ms−10.32\,m{s^{ - 1}}

Explanation

Solution

Oscillating motion is a sort of simple harmonic motion. In this type of motion, a particle always follows the same route, accelerating towards a fixed point. To answer the question of what is the maximum velocity of a simple pendulum, we will use the formula for determining the velocity of a simple pendulum experiencing SHM to obtain the required answer.

Formula used:
vmax⁡=ωa{v_{\max }} = \omega a
Where aa is the amplitude and ω\omega is the angular velocity.

Complete step by step answer:
We are given the time and amplitude in the given question, and we will obtain the desired answer for the query by inserting the values in a suitable formula.Given: Time period T=2 secT = 2\,sec and Amplitude of pendulum A=50 mm=0.05 mA = 50\,mm = 0.05\,m. We know that the velocity of a simple pendulum undergoing SHM is given by;
v=ωa2−u2v = \omega \sqrt {{a^2} - {u^2}}

The velocity is at its maximum in the equilibrium position, while the acceleration (a)\left( a \right) is zero. Simple harmonic motion is characterised by a variable acceleration that is proportionate to the displacement from the equilibrium point and is always directed toward it.When, u=0, v=vmax⁡u = 0,\,v = {v_{\max }}
So, vmax⁡=ωa{v_{\max }} = \omega a
[Where aa is the amplitude and ω\omega is the angular velocity]
vmax⁡=2πT×a ⇒vmax⁡=2π2×0.05 ∴vmax⁡=0.16 ms−1{v_{\max }} = \dfrac{{2\pi }}{T} \times a \\\ \Rightarrow {v_{\max }} = \dfrac{{2\pi }}{2} \times 0.05 \\\ \therefore {v_{\max }} = 0.16\,m{s^{ - 1}}
Therefore, the maximum velocity will be 0.16 ms−10.16\,m{s^{ - 1}}.

Hence, the correct option is B.

Note: In simple harmonic motion, the system's acceleration, and hence the net force, is proportionate to the displacement and acts in the opposite direction. Also, one important point to remember about SHM is that, for one thing, the time TT and frequency ff of a simple harmonic oscillator are unaffected by amplitude.