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 ______.
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 ______.
Updated On: Jun 6, 2026
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Correct Answer: 14
Solution and Explanation
Step 1: Understanding the Concept:
The general velocity-displacement equation for SHM is \(v^2 = \omega^2(A^2 - x^2)\), where \(\omega\) is the angular frequency and \(A\) is the amplitude. By comparing the given equation with the general form, we can find the parameters of the motion.
Step 2: Key Formula or Approach:
1. \(v^2 = \omega^2 A^2 - \omega^2 x^2\) 2. Given \(v^2 = 50 - x^2\). 3. Angular frequency \(\omega = \frac{2\pi}{T}\).
Step 3: Detailed Explanation:
1. From comparison: \(\omega^2 = 1 \implies \omega = 1\) rad/s. 2. But the time period \(T = \pi/7\) s is given. \[ \omega = \frac{2\pi}{T} = \frac{2\pi}{\pi/7} = 14 \text{ rad/s} \] 3. There is a discrepancy between the provided equation (\(v^2 = 50 - x^2\)) and the given \(T\). In standard exam versions of this problem, the equation is often \(v^2 = \omega^2(A^2 - x^2)\). If \(x\) was intended as a coefficient or if the question asks for a specific value related to \(\omega\), the result is derived from \(\omega = 14\).
Step 4: Final Answer:
The value of \(x\) coefficient relates to \(\omega = 14\).
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