Question

A small object is attached to a horizontal spring, pushed to position x=−A, and released. The object in the resulting simple harmonic motion oscillates with period T.

How long does it take for the object to travel a total distance of 3A?

O 1/2 T
O 3/4 T
O 5/4 T
O T
O 1/4T

Answer 1

The time it takes for an object performing simple harmonic motion (SHM) to travel a distance of 3A is 3/4T. This is because a full cycle of SHM is two times the amplitude A and takes one period T. Half of this cycle plus one-quarter of the cycle equals 3/4T. Hence, the second option is correct.

Step-by-step explanation:

The student's question revolves around calculating the time it takes for an object in simple harmonic motion (SHM) to travel a total distance of 3A, where A is the amplitude and T is the period of the motion. In SHM, an object oscillates symmetrically on either side of the equilibrium position, covering a distance of 2A in one full cycle (T).

Here is a step-by-step explanation:

1. When the object is released from the position x = -A, it moves towards the equilibrium (0) covering a distance of A.
2. It then continues to the opposite side to x = A, traveling another distance of A (making a total of 2A).
3. Now, the object must move back towards the equilibrium position, covering the remaining A to complete the total distance of 3A. This final leg represents half the distance of a full cycle.

The first half-cycle (from -A to A) takes 1/2 T, and the additional 1/2A segment takes an additional 1/4 T, as it is half of a half-cycle. When we add up these times, we get a total time of 3/4 T to travel a distance of 3A.