Question

If the equation for the displacement of a particle executing simple harmonic motion is $x = 3 \sin \left( \frac{2\pi t}{18} + \frac{\pi}{6} \right)$ cm, then the distance travelled by the particle in a time of 36 s is
(Time $t$ is in seconds)

• A. 24 cm
• B. 12 cm
• C. 18 cm
• D. 15 cm

Hint:

In simple harmonic motion, the distance travelled in one complete oscillation is always $4A$, regardless of the phase constant.

Answer 1

The Correct Option is A

The displacement equation is \( x = 3 \sin \left( \frac{2\pi t}{18} + \frac{\pi}{6} \right) \) cm. The general form of SHM is \( x = A \sin(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. Here, \( A = 3 \) cm, \( \omega = \frac{2\pi}{18} = \frac{\pi}{9} \) rad/s, and \( \phi = \frac{\pi}{6} \).

The period \( T \) of the motion is given by \( T = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{\pi}{9}} = 18 \) s. In 36 s, the number of complete oscillations is \( \frac{36}{18} = 2 \).

In one complete oscillation, the particle travels a distance equal to 4 times the amplitude (from \(-A\) to \(+A\) and back): \( 4A = 4 \times 3 = 12 \) cm. For 2 oscillations, the distance is \( 2 \times 12 = 24 \) cm. The phase does not affect the total distance travelled in complete cycles.

Thus, the distance travelled in 36 s is 24 cm.