Question

A particle executes S.H.M. from extreme position and covers a distance equal to half of its amplitude in 1 s. Determine the time period of motion

• A. 6 seconds
• B. 3 seconds
• C. 1 second
• D. 2 seconds

Answer 1

Answer: (A)

6 seconds

Answer 2

The displacement $x(t)$ of a particle executing Simple Harmonic Motion (SHM) from the mean position, when starting from an extreme position, is given by the equation: $x(t) = a \cos(\omega t)$ where $a$ is the amplitude and $\omega$ is the angular frequency. At $t=0$, the particle is at the extreme position, so $x(0) = a$. The problem states that the particle covers a distance equal to half of its amplitude ($a/2$) in $1$ second. This implies that after $1$ second, the particle's displacement from the mean position is $x(1) = a - a/2 = a/2$ (as it moves towards the mean position). Substituting $t=1$s and $x(1) = a/2$ into the equation of motion: $\frac{a}{2} = a \cos(\omega \times 1)$ $\frac{1}{2} = \cos(\omega)$ The smallest positive value for $\omega$ that satisfies this equation is $\omega = \frac{\pi}{3}$ radians. The relationship between angular frequency $\omega$ and the time period $T$ is: $\omega = \frac{2\pi}{T}$ Substituting the value of $\omega$: $\frac{\pi}{3} = \frac{2\pi}{T}$ Solving for $T$: $T = \frac{2\pi}{\pi/3} = 2\pi \times \frac{3}{\pi} = 6$ Therefore, the time period of the motion is $6$ seconds.