Question

A body is vibrating in simple harmonic motion with amplitude of $0.06$ m and frequency of $15$ Hz. The maximum velocity and acceleration of the body is:

• A. 9.80 m/s and $ 9.03\,\times {{10}^{2}}\,m/{{s}^{2}} $
• B. 8.90 m/s and $ 8.21\times {{10}^{2}}\,m/{{s}^{2}} $
• C. 6.82 m/s and $ 7.62\times {{10}^{2}}\,m/{{s}^{2}} $
• D. 5.65 m/s and $ 5.32\,\times {{10}^{2}}\,m/{{s}^{2}} $

Answer 1

Answer: (D)

When particle passes through its equilibrium position, then the velocity is maximum and acceleration is maximum at extreme position.
The velocity of a particle in SHM is given by
$ u=\omega \sqrt{{{a}^{2}}-{{y}^{2}}} $
where $ \omega $ is angular velocity, a the amplitude and y the displacement.
When the particle passes through its equilibrium position, then the velocity is maximum.
$ {{u}_{\max }}=a\omega $ ....(i)
Similarly maximum acceleration at maximum displacement
$ {{\alpha }_{\max }}={{\omega }^{2}}a $ ....(ii)
Given $ a=0.06\,m,\omega =2\pi f=2\pi \times 15 $
Putting these values in Eqs. (i) and (ii), we get
$ {{u}_{\max }}=0.06\times 2\pi \times 15 $
$ =0.06\times 2\times 3.14\times 15 $
$ =5.65\,m/s $
$ {{\alpha }_{\max }}=5.32\times {{10}^{2}}m/{{s}^{2}} $