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 $a=0.06m,\omega =2\pi f=2\pi \times 15$
Putting these values in Eqs. (i) and (ii), we get
$u_{\max }=0.06\times 2\pi \times 15$
$=0.06\times 2\times 3.14\times 15$
$=5.65m/s$
${\alpha }_{\max }=5.32\times {10}^{2}m/s^2$