JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2026
- Medium
- Numerical
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 \(\dfrac{x}{7}\) s, the value of \(x\) is _____.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 44
Step-by-step Solution
Detailed explanation
The given equation is \(v^2 = 50 - x^2\)
Differentiating with respect to \(x\):
\(2v \dfrac{dv}{dx} = -2x\)
\(v \dfrac{dv}{dx} = -x\)
Since acceleration \(a = v \dfrac{dv}{dx}\), we get:
\(a = -x\)
Comparing this with the standard equation of SHM \(a = -\omega^2 x\), we get:
\(\omega^2 = 1 \Rightarrow \omega = 1\) rad/s
The time period of the motion is given by:
\(T = \dfrac{2\pi}{\omega} = 2\pi\)
Taking \(\pi = \dfrac{22}{7}\), we get:
\(T = 2 \times \dfrac{22}{7} = \dfrac{44}{7}\) s
Given that the time period is \(\dfrac{x}{7}\) s, comparing the two gives:
\(x = 44\)
Answer: \(44\)
Differentiating with respect to \(x\):
\(2v \dfrac{dv}{dx} = -2x\)
\(v \dfrac{dv}{dx} = -x\)
Since acceleration \(a = v \dfrac{dv}{dx}\), we get:
\(a = -x\)
Comparing this with the standard equation of SHM \(a = -\omega^2 x\), we get:
\(\omega^2 = 1 \Rightarrow \omega = 1\) rad/s
The time period of the motion is given by:
\(T = \dfrac{2\pi}{\omega} = 2\pi\)
Taking \(\pi = \dfrac{22}{7}\), we get:
\(T = 2 \times \dfrac{22}{7} = \dfrac{44}{7}\) s
Given that the time period is \(\dfrac{x}{7}\) s, comparing the two gives:
\(x = 44\)
Answer: \(44\)
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