ExamBro
ExamBro
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.
Verified Solution

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: \(44\)
Same subject
Explore more questions on app