Question

The velocity of a particle executing simple harmonic motion along x-axis is described as \(v^2 = 50 - x^2\), where \(x\) represents displacement. If the time period of motion is \(\pi/7\) s, the value of \(x\) is ______.

Answer 1

Correct Answer: 14

48. Step 1: Use the standard SHM equation
[4pt] For simple harmonic motion, \[ v^2=\omega^2(A^2-x^2) \] where: \[ \omega = \text{angular frequency}, \qquad A = \text{amplitude} \] Step 2: Compare with the given equation
[4pt] Given: \[ v^2=50-x^2 \] Comparing with \[ v^2=\omega^2(A^2-x^2) \] we get: \[ \omega^2=1 \] Hence, \[ \omega=1\ \text{rad/s} \] Also, \[ A^2=50 \] \[ A=\sqrt{50} \] Step 3: Find the time period
[4pt] The time period of SHM is: \[ T=\frac{2\pi}{\omega} \] Substituting \(\omega=1\): \[ T=2\pi \] Using \[ \pi=\frac{22}{7} \] \[ T=2\times\frac{22}{7} \] \[ T=\frac{44}{7}\text{ s} \] Step 4: Compare with given time period
[4pt] Given: \[ T=\frac{x}{7}\text{ s} \] So, \[ \frac{x}{7}=\frac{44}{7} \] Multiplying both sides by \(7\): \[ x=44 \] Final Answer: \[ \boxed{x=44} \]