Question

A $20\,g$ particle moves in SHM with a frequency of $3$ oscillations per second and amplitude of $5\,cm$ . Through what total distance does the particle move during one oscillation? What is its Average speed?
A. $20\,cm,20\,cm\,{\sec ^{ - 1}}$
B. $40\,cm,40\,cm\,{\sec ^{ - 1}}$
C. $40\,cm,60\,cm\,{\sec ^{ - 1}}$
D. $20\,cm,60\,cm\,{\sec ^{ - 1}}$

Answer 1

Hint: In order to solve this question, we will use the general SHM equation of motion where we know it’s mass, frequency and amplitude and will solve for its displacement for one oscillation. And we know the time period of a particle is related to frequency as $T = \dfrac{1}{f}$ .

Complete step by step answer:
We have given that the amplitude of the particle is $5\,cm$ and it’s performing $3$ oscillations per second having mass of $20\,g$. So, the frequency of the particles is given by: $f = 3Hz$. And the angular frequency of a particle performing SHM is given by:
$\omega = 2\pi f$
$\omega = 6\pi \,radian\,{\sec ^{ - 1}}$
Which means, in one complete oscillation particle moves from its mean position to one of extreme position and then returns to its mean position and later again moves to another extreme position and finally comes back to its original mean position.

In one complete oscillation, the particle covers four maximum displacements which is equal to its amplitude of $5\,cm$.Hence, total distance covered is $4 \times 5 = 20\,cm$
And now for average speed we know that,
$\text{speed} = \dfrac{\text{distance}}{\text{time}}$
$\Rightarrow \text{speed} = 20 \times 3$
$\therefore \text{speed} = 60\,cm\,{\sec ^{ - 1}}$

Hence, the correct answer is option D.

Note:In SHM , particles displace from its mean position to its extreme position and this displacement is maximum in its whole motion and this displacement is known as amplitude of the particle and total time of the particle is time taken to complete its one oscillation.