Question

The mean velocity of a particle performing S.H.M. with a time period of $ 0.6s $ and amplitude of $ 10cm $ averaged over a time interval during which it travels a distance of $ 5cm, $ starting from the extreme position is
(A) $ 0.5m/s $
(B) $ 0.7m/s $
(C) $ 0.3m/s $
(D) $ 1.04m/s $

Answer 1

Hint
Simple harmonic motion is the example of the simplest form of periodic motion. An oscillatory motion is assumed to be simple harmonic when the displacement of the particle from the origin varies with time. We know that the velocity of a particle is the displacement divided by time.
 $\Rightarrow x = A\cos \omega t $ (Where $ x $ stands for the displacement of the particle, $ A $ stands for the amplitude of the oscillation. $ \omega $ stands for the frequency of the oscillation and $ t $ stands for the time)
 $\Rightarrow \omega = \dfrac{{2\pi }}{T} $ (Where, $ \omega $ stands for the frequency of the oscillation and $ T $ stands for the time period of oscillation)
 $\Rightarrow x = \dfrac{v}{t} $ (Where $ x $ stands for the displacement, $ v $ stands for the velocity and $ t $ stands for the time)

Complete step by step answer
The time period of oscillation is given by,
 $\Rightarrow T = 0.6s $
The amplitude of the oscillation is given as,
 $\Rightarrow A = 10cm $
The displacement from the mean position when the particle travels $ 5cm $ from the extreme position can be written as,
 $\Rightarrow x = 10 - 5 = 5cm $
Let $ t $ be the time taken for displacement,
Then the expression for displacement can be written as,
 $\Rightarrow x = A\cos \omega t $
Putting the values of $ x $ and $ A $ in the above expression,
 $\Rightarrow 5 = 10\cos \omega t $
 $\Rightarrow \dfrac{5}{{10}} = \cos \omega t $
 $\Rightarrow \dfrac{1}{2} = \cos \omega t $
From this we get,
 $\Rightarrow \cos \omega t = \dfrac{1}{2} $
 $\Rightarrow \omega t = \dfrac{\pi }{3} $
We can write,
 $\Rightarrow \omega = \dfrac{{2\pi }}{T} $
Substituting this value of $ \omega $ in the above equation, we get
 $\Rightarrow \left( {\dfrac{{2\pi }}{T}} \right)t = \dfrac{\pi }{3} $
From the equation, we get
 $\Rightarrow t = \dfrac{T}{6} $
The time period, $ T = 0.6s $
Substituting the value of time period in the above equation, we get
 $\Rightarrow t = \dfrac{{0.6}}{6} = 0.1s $
We know that, the displacement is,
 $\Rightarrow x = 5cm = 5 \times {10^{ - 2}}m $
The velocity of the particle will be,
 $\Rightarrow v = \dfrac{x}{t} = \dfrac{{5 \times {{10}^{ - 2}}}}{{0.1}} = 0.5m/s $
The answer is: Option (A): $ 0.5m/s $ .

Note
In this question we use the trigonometric identity $ \cos \dfrac{\pi }{3} = \dfrac{1}{2} $ . The frequency of simple harmonic motion is defined as the number oscillations per unit time. The time period of an oscillation is defined as the time taken for the execution of one complete oscillation. We use the relation between frequency and time period to find the time taken for the oscillation.