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Question: The displacement y (in cm) produced by a simple harmonic wave is \[y = \dfrac{{10}}{\pi }\sin \]\[\l...

The displacement y (in cm) produced by a simple harmonic wave is y = \dfrac{{10}}{\pi }\sin $$$$\left( {200\pi t - \dfrac{{\pi x}}{{17}}} \right) periodic time and maximum velocity of the particle in the medium will respectively be-
(A) 10−3sec⁡and 330m/sec⁡{10^{ - 3}}\sec and \, 330m/\sec
(B) 10−4sec⁡and 20m/sec⁡{10^{ - 4}}\sec and \, 20m/\sec
(C) 10−2sec⁡and 20m/sec⁡{10^{ - 2}}\sec and \, 20m/\sec
(D) 10−2sec⁡and 200m/sec⁡{10^{ - 2}}\sec and \, 200m/\sec

Explanation

Solution

In order to solve this problem,we are going to apply the concept of simple harmonic motion(SHM). It is a back and forth movement along with an equilibrium, or center position, so that the maximum displacement on one side of that position is equal to the maximum displacement on the other side position. The time period of a particular interval of a complete vibration is similar.

Complete step by step answer:
The force which is a cause for the motion is always pointed toward the equilibrium position and is directly proportional to the distance from that position. The force is given by formula, F=−kxF = - kx, where F is the force, x is the displacement, and k is a constant. The force equation is known as the Hooke’s law.There is a very similar connection between simple harmonic motion(SHM) and periodic waves, in most of the periodic waves types, the particles in that wave medium experience the simple harmonic motion. In a longitudinal wave type for example such as a sound wave propagation, the particles in that wave oscillates along the direction of harmonic motion of the wave.
By using formula for calculating periodic time in seconds and velocity we can get the solution,
So, comparing with standard equation y=Asin⁡(ωt−kx)y = A\sin \left( {\omega t - kx} \right) we get,
As we have,
ω=200π\omega = 200\pi k=π17k = \dfrac{\pi }{{17}}
We get, 2πT=200π\dfrac{{2\pi }}{T} = 200\pi
T=10−2sec⁡T = {10^{ - 2}}\sec
And Vmax⁡=Aω=(10π)100×200π{V_{\max }} = A\omega = \dfrac{{\left( {\dfrac{{10}}{\pi }} \right)}}{{100}} \times 200\pi
∴Vmax⁡=20m/sec⁡\therefore{V_{\max }} = 20m/\sec

Therefore correct answer is option C.

Note: Simple Harmonic Motion Defines that it is a harmonic motion in which the object moves to and fro,back and forth along the line of propagation. For example, when a person swings that object, it moves to and fro along the direction of line. These movements are known as oscillations. Oscillations of a pendulum are one of the examples of simple harmonic motion.