Question

The velocity of a particle executing SHM varies with displacement (\(x\)) as \(4v^2 = 50-x^2\).
The time period of oscillations is \(x/7\) s.
The value of \(x\) is (Take \(π = 22/7\))

• A. \(82\)
• B. \(84\)
• C. \(88\)
• D. \(90\)

Hint:

Compare with \(v^2=\omega^2(A^2-x^2)\) to find \(\omega\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In SHM, \(v^2 = \omega^2(A^2 - x^2)\).

Step 2: Compare:
The given relation is \(v^2 = \frac14(50 - x^2)\). So \(\omega^2 = \frac14\), giving \(\omega = \frac12\) rad/s, and \(A^2 = 50\).

Step 3: Time period:
\[ T = \frac{2\pi}{\omega} = 4\pi = 4\times\frac{22}{7} = \frac{88}{7}\ \text{s} \]
The problem says \(T = \frac{x}{7}\), so \(x = 88\).

Final Answer:
The value of \(x\) is \(88\), option (C). \[ \boxed{88} \]