Question

The time period of a particle executing SHM is 2s. After what time interval from t=0, will its displacement be half of the amplitude:
A. (1/4)secs
B. (1/3) secs
C. (1/2)secs
D. 1 sec

Answer 1

Hint: Recall that simple harmonic motion is a time varying motion in which amplitude changes with time. Use the mathematical expression for simple harmonic motion to find the displacement. Also take the value of \[\omega \] as \[\dfrac{2\pi }{T}\] .

Complete step-by-step answer:
Periodic motion is a kind of motion that repeats at a regular interval of time. Simple harmonic motion is a type of periodic motion. When a particle is displaced from its mean position, there is a restoring force acting on it which tends to bring it back to its mean position in the case of a simple harmonic motion. This restoring force is proportional to the displacement of the particle, but in the opposite direction. All periodic motions are not simple harmonic motion. The examples of simple harmonic motion are a simple pendulum and a mass attached to a spring with one end fixed to a point.
So we can write the relation as
 \[\mathbf{F}\alpha \text{ }\mathbf{x}\] or
 \[\mathbf{F}=-k\mathbf{x}\]
Where F- restoring force
             k- constant
             x- displacement vector.
Remember that the restoring force always tries to bring back the particle to the mean position. In the case of a mass attached spring k is its spring constant.
It is a back and forth motion, in which the amplitude varies with time, the motion is sinusoidal with time. Time period is the total time taken to complete one cycle.
The displacement from the mean position at a given time t is given as
 \[y=A\cos (\omega t)\] ……………(1)
where A -amplitude or maximum displacement of the particle
              \[\omega \] - angular frequency
We have to find the time at which the amplitude becomes half.
Given data are
Time period of the pendulum, T =2s
Amplitude or displacement of the pendulum at time t, \[y=\dfrac{A}{2}\]
Using equation (1), we can write
 \[\dfrac{A}{2}=A\cos (\omega t)\] …………(2)
But \[\omega =\dfrac{2\pi }{T}\] . Substituting the value of \[\omega \] in equation (2)
 \[\dfrac{A}{2}=A\cos \left( \dfrac{2\pi }{T}t \right)\]
By simplifying, we get
 \[\dfrac{1}{2}=\cos \left( \dfrac{2\pi }{2}t \right)\]
 \[\dfrac{1}{2}=\cos \left( \pi t \right)\] …………..(3)
But \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\] becomes
 \[\cos \dfrac{\pi }{3}=\cos (\pi t)\] .
 \[\dfrac{\pi }{3}=\pi t\] or
 \[t=\dfrac{1}{3}s\] .
This implies that when at the time \[t=\dfrac{1}{3}s\] , the displacement of the wave will become half of that of the amplitude.
So option C is the correct answer.

Note: Remember this is a periodic process, so the amplitude of the motion changes with time. Here the amplitude changes from –A ( \[\theta ={{180}^{0}}\] ) to A( \[\theta ={{0}^{0}}\] ). At \[\theta ={{90}^{o}}\text{ }or\text{ }{{270}^{0}}\] , the amplitude is zero. The total angle after one complete cycle will be \[{{360}^{0}}\] .