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: Plan:
Differentiate the relation to get the acceleration.

Step 2: Steps:
$4v^2 = 50 - x^2$. Differentiate with respect to time: $8v\frac{dv}{dt} = -2x\frac{dx}{dt} = -2xv$, so $a = \frac{dv}{dt} = -\frac{x}{4}$.
For SHM, $a = -\omega^2x$, so $\omega^2 = \frac14$ and $T = \frac{2\pi}{1/2} = 4\pi = \frac{88}{7}$ s. Hence $x = 88$.

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