Question

The distance travelled by a particle executing linear S.H.M. from its mean position in 2 s is equal to \( \frac{1}{\sqrt{2}} \) times its amplitude. Then its time period in seconds is

• A. \(10 \)
• B. \(8 \)
• C. \(9 \)
• D. \(12 \)
• E. \(16 \)

Hint:

Use \( x = A\sin(\omega t) \) for SHM position.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The position (displacement) of a particle executing Simple Harmonic Motion (S.H.M.) starting from its mean position is described by a sine function of time. We can use this equation to relate position, amplitude, time, and time period.
Step 2: Key Formula or Approach:
The equation for the displacement `x` of a particle in S.H.M. starting from the mean position is: \[ x(t) = A \sin(\omega t) \] where A is the amplitude, \(\omega\) is the angular frequency, and `t` is the time. The angular frequency is related to the time period T by \(\omega = \frac{2\pi}{T}\). We are given the condition at \(t=2\)s to find T.
Step 3: Detailed Explanation:
We are given that at time \(t = 2\)s, the distance from the mean position is \(x = \frac{A}{\sqrt{2}}\). Substitute these values into the S.H.M. equation: \[ \frac{A}{\sqrt{2}} = A \sin(\omega \cdot 2) \] The amplitude A cancels out: \[ \frac{1}{\sqrt{2}} = \sin(2\omega) \] The principal value for the angle whose sine is \(\frac{1}{\sqrt{2}}\) is \(\frac{\pi}{4}\). So, we have: \[ 2\omega = \frac{\pi}{4} \] \[ \omega = \frac{\pi}{8} \] Now, we use the relationship between angular frequency and time period: \[ \omega = \frac{2\pi}{T} \] \[ \frac{\pi}{8} = \frac{2\pi}{T} \] The \(\pi\) cancels out: \[ \frac{1}{8} = \frac{2}{T} \] Solving for T: \[ T = 2 \times 8 = 16 \text{ seconds} \] Step 4: Final Answer:
The time period is 16 seconds.