Question

For a particle performing linear S.H.M., its average speed over one oscillation is (a = amplitude of S.H.M., n = frequency of oscillation)

• A. $2\, an$
• B. $4 \,an$
• C. $6 \,an$
• D. $8\,an$

Answer 1

Answer: (B)

Given, amplitude of $SHM = a$
Frequency of oscillation $=n$
Distance travelled in one oscillation,
$=4 \times \text { amplitude of } SHM =4 a$
We know that,
$\text { Velocity } =\frac{\text { Distance travelled in one oscillation }}{\text { Time period }} $
$v =\frac{4 a}{T}$
$\therefore \,\,\,\, v=4 a n\,\,\,\,\,\,\,\,\,$ $\left[\because\right.$ Frequency, $\left.n=\frac{1}{T}\right]$