Question

A body of mass 1/4 kg is in S.H.M and its displacement is given by the relation \[y = 0.05\sin \left( {20t + \dfrac{\pi }{2}} \right)\,m\]. If t is in seconds, the maximum force acting on the particle is:
A. 5 N
B. 2.5 N
C. 10 N
D. 0.25 N

Answer 1

Hint: Compare the given equation with the equation of the transverse wave and determine the values of amplitude and angular frequency. Use the formula for particles performing S.H.M. to determine the maximum acceleration of the particle. Then you can use the formula for force using Newton’s second law.

Formula used:
The force F is given by Newton’s second law of motion is,
\[F = ma\]
Here, m is the mass and a is the acceleration.


Complete step by step answer:
We have the ideal equation of transverse wave is,
\[y = A\sin \left( {\omega t - kx + \phi } \right)\]

Here, A is the amplitude of oscillations, \[\omega \] is the angular frequency, x is the horizontal distance, k is the force constant and \[\phi \] is the phase constant.

We have given the equation of the transverse wave, \[y = 0.05\sin \left( {20t + \dfrac{\pi }{2}} \right)\,m\].

Now, if we compare the above two equations, we will get the amplitude and angular frequency of the particle. Therefore,
\[A = 0.05\,m\] and \[\omega = 20\,rad/s\].

We know that, the acceleration of the particle performing simple harmonic motion is given by,
\[a = {\omega ^2}x\]

Here, x is the displacement of the particle. We know that the particle has maximum acceleration at extreme positions that are at \[x = + A\,\,{\text{and}}\,\,x = - A\]. Therefore, we can substitute A for x in the above equation.
\[{a_{\max }} = {\omega ^2}A\]

Substitute \[\omega = 20\,rad/s\] and \[A = 0.05\,m\] in the above equation.
\[{a_{\max }} = {\left( {20} \right)^2}\left( {0.05} \right)\]
\[ \Rightarrow {a_{\max }} = 20\,m/{s^2}\]

We have the maximum force acting on the particle of mass m is,
\[F = m{a_{\max }}\]

Now, we can substitute \[m = \dfrac{1}{4}kg\] and \[{a_{\max }} = 20\,m/{s^2}\] in the above equation.
\[{F_{\max }} = \left( {\dfrac{1}{4}} \right)\left( {20} \right)\]
\[ \Rightarrow {F_{\max }} = 5\,N\]

So, the correct answer is “Option A”.

Note:
To solve such types of questions, students should always compare the given equation of the transverse wave with the ideal equation of the transverse wave. On comparing you will get the values of angular velocity, wave number, wavelength and phase constant of the wave. You should always remember the maximum values of acceleration and velocity of the particle performing S.H.M.