Question

The period of oscillation of a particle in simple harmonic motion is 4 s, and its amplitude is 4 cm. Then the distance of the particle in \( \frac{1}{3} \) seconds after passing the mean position is:

• A. 1.5 cm
• B. 2 cm
• C. 2.3 cm
• D. 2.5 cm

Hint:

For simple harmonic motion, the displacement can be calculated using the equation \( x(t) = A \cos(\omega t) \), where \( A \) is the amplitude and \( \omega \) is the angular frequency.

Answer 1

The Correct Option is B

The displacement \( x(t) \) of a particle undergoing simple harmonic motion is given by the equation:

\[ x(t) = A \cos(\omega t) \] where:
- \( A \) is the amplitude of the oscillation,
- \( \omega = \frac{2\pi}{T} \) is the angular frequency, and
- \( T \) is the time period of the motion.

Here, we are given that:
- \( A = 4 \) cm
- \( T = 4 \) s
- We need to find the displacement after \( t = \frac{1}{3} \) seconds.

Step 1: Calculate the angular frequency \( \omega \)
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{rad/s} \]

Step 2: Substitute values into the displacement equation
\[ x(t) = 4 \cos\left(\frac{\pi}{2} \times \frac{1}{3}\right) \] \[ x(t) = 4 \cos\left(\frac{\pi}{6}\right) \] \[ x(t) = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \, \text{cm} \]

Final Answer:
Therefore, the distance of the particle after \( \frac{1}{3} \) seconds is approximately:
\[ \boxed{2 \, \text{cm}} \]