Question

4. A 20 g particle executing SHM with amplitude
of 5 cm. The total distance covered during one
oscillation is
a. 10 cm
b. 15 cm
c. 20 cm
d. 25 cm​

Answer 1

The answer is c)20 cm.

We know, distance covered in one oscillations in SHM = 4 × amplitude .

For more details,see attachments.

Regards

Kshitij

Answer 2

Answer:

Given:

Particle covers an amplitude of 5 cm.

Mass of particle = 20 g

To find:

Total distance travelled in a single Oscillation.

Concept:

Let's consider that the particle starts from a Mean Position. The particle moves to one Extreme position ( hence travels one amplitude). Again it returns back to mean position ( hence travels 2nd amplitude).

Now from the mean position , it again goes to the opposite Extreme position ( hence travels 3rd Amplitude ) . Finally comes back to mean position ( 4th amplitude).

Thus the particle travels (4 × Amplitude) while completing a single Oscillation.

Calculation:

$\therefore \: total \: distance$

$= 4 \times amplitude$

$= 4 \times 5$

$= 20 \: cm$

So final answer is :

$\boxed{ \boxed{ \orange{ \sf{ \large{total \: distance \: = 20cm}}}}}$