Question

A particle executes simple harmonic motion between \( x = -A \) and \( x = +A \). If the time taken by the particle to go from \( x = 0 \) to \( \frac{A}{2} \) is 2 s, then the time taken by the particle in going from \( x = \frac{A}{2} \) to \( A \) is:

• A. 3 s
• B. 2 s
• C. 1.5 s
• D. 4 s

Hint:

In SHM, the time taken to travel between points is not uniform and depends on the amplitude and angular frequency.

Answer 1

The Correct Option is D

Step 1: {Using the standard equation of SHM}
\[ \frac{A}{2} = A \sin(\omega t_1) \] \[ \omega t_1 = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \quad {(i)} \] Step 2: {Calculating total time to reach \( A \)}
\[ A = A \sin \omega (t_1 + t_2) \] \[ \omega (t_1 + t_2) = \sin^{-1}(1) = \frac{\pi}{2} \] \[ \omega t_2 = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \quad {(ii)} \] Step 3: {Finding the ratio of times}
\[ \frac{t_1}{t_2} = \frac{1}{2} \Rightarrow t_2 = 2t_1 = 2 \times 2 = 4 \, {s} \] Thus, the time taken to go from \( \frac{A}{2} \) to \( A \) is 4 s.

Answer 2

Step-by-step Solution:

Step 1: Understand the motion of a particle in SHM
In simple harmonic motion (SHM), the displacement \( x \) at any time \( t \) can be described as:
\( x(t) = A \sin(\omega t) \) or \( x(t) = A \cos(\omega t) \), depending on the starting position.
Here, we assume the particle starts from the mean position \( x = 0 \), so the equation is:
\( x = A \sin(\omega t) \)

Step 2: Given condition and find angular frequency
We are given that the particle takes 2 seconds to move from \( x = 0 \) to \( x = \dfrac{A}{2} \).
Let’s find the corresponding angle \( \theta \) for this position:
If \( x = A \sin(\omega t) = \dfrac{A}{2} \), then:
\( \sin(\omega t) = \dfrac{1}{2} \Rightarrow \omega t = \dfrac{\pi}{6} \)

Now, using \( t = 2 \, \text{s} \):
\( \omega \cdot 2 = \dfrac{\pi}{6} \Rightarrow \omega = \dfrac{\pi}{12} \)

Step 3: Find time to go from \( x = \dfrac{A}{2} \) to \( x = A \)
Now find the time taken to move from \( x = \dfrac{A}{2} \) to \( x = A \):
At \( x = A \), \( \sin(\omega t) = 1 \Rightarrow \omega t = \dfrac{\pi}{2} \)

So the time taken from \( x = 0 \) to \( x = A \) is:
\( t = \dfrac{\pi}{2} \div \omega = \dfrac{\pi}{2} \cdot \dfrac{12}{\pi} = 6 \, \text{s} \)

Therefore, time taken from \( x = \dfrac{A}{2} \) to \( x = A \) is:
\( 6 \, \text{s} - 2 \, \text{s} = 4 \, \text{s} \)

Final Answer:
The time taken by the particle to move from \( x = \dfrac{A}{2} \) to \( x = A \) is 4 seconds.