Question

A particle ís executing simple harmonic motion with amplitude of $0.1\text{m}$. At a certain instant when its displacement is $0.02\text{m}$, its acceleration is $0.5{\text{ms}}^{-2}$. The maximum velocity of the particle is (in ${\text{ms}}^{-1}$)

• A. $0.01$
• B. $0.05$
• C. $0.5$
• D. $0.25$

Answer 1

The correct option is C $0.5$

Acceleration, $a=0.5{\text{ms}}^{-2}$
Amplitude, $A=0.1\text{m}$
Displacement, $y=0.02\text{m}$
Using the formula of maximum acceleration $a={\omega }^{2}y$
or $0.5={\omega }^2\times 0.02$
or ${\omega }^2=\frac{0.5}{0.02}=25$
So, $\omega =5$
Now, maximum velocity is
$v=A\omega =0.1\times 5=0.5{\text{ms}}^{-1}$