Question

A particle performs SHM, having a speed of 6 cm/sec at the mean position and an amplitude of 4 cm. Find the position of the particle from the mean position when its velocity is 2 cm/sec.

• A. \(2 \, \text{cm}\)
• B. \(3 \, \text{cm}\)
• C. \(4 \, \text{cm}\)
• D. \(1 \, \text{cm}\)

Hint:

For SHM problems, remember that velocity depends on the position, and use the formula to relate them.

Answer 1

The Correct Option is C

To determine the position of the particle from the mean position when its velocity is \(2 \, \text{cm/sec}\), we can use the principles of Simple Harmonic Motion (SHM). Given: Maximum speed at mean position, \(v_{\text{max}} = 6 \, \text{cm/sec}\) Amplitude of SHM, \(A = 4 \, \text{cm}\) Velocity at position \(x\), \(v = 2 \, \text{cm/sec}\) Step 1: Relate Maximum Speed to Angular Frequency In SHM, the maximum speed \(v_{\text{max}}\) is related to the angular frequency \(\omega\) and amplitude \(A\) by the equation: \[ v_{\text{max}} = \omega A \] Solving for \(\omega\): \[ \omega = \frac{v_{\text{max}}}{A} = \frac{6 \, \text{cm/sec}}{4 \, \text{cm}} = 1.5 \, \text{sec}^{-1} \] Step 2: Use the Velocity-Position Relationship in SHM The velocity \(v\) of a particle in SHM at position \(x\) is given by: \[ v = \omega \sqrt{A^2 - x^2} \] Solving for \(x\): \[ v = \omega \sqrt{A^2 - x^2}
\Rightarrow \sqrt{A^2 - x^2} = \frac{v}{\omega}
\Rightarrow A^2 - x^2 = \left( \frac{v}{\omega} \right)^2
\Rightarrow x^2 = A^2 - \left( \frac{v}{\omega} \right)^2
\Rightarrow x = \sqrt{A^2 - \left( \frac{v}{\omega} \right)^2} \] Step 3: Substitute the Known Values Substitute \(A = 4 \, \text{cm}\), \(v = 2 \, \text{cm/sec}\), and \(\omega = 1.5 \, \text{sec}^{-1}\) into the equation: \[ x = \sqrt{4^2 - \left( \frac{2}{1.5} \right)^2}
x = \sqrt{16 - \left( \frac{4}{2.25} \right)}
x = \sqrt{16 - \frac{16}{9}}
x = \sqrt{\frac{144}{9} - \frac{16}{9}}
x = \sqrt{\frac{128}{9}}
x = \frac{\sqrt{128}}{3}
\]\[ x = \frac{8 \sqrt{2}}{3} \approx 3.77 \, \text{cm} \] Rounding to the nearest whole number: \[ x \approx 4 \, \text{cm} \] Conclusion: The position of the particle from the mean position when its velocity is \(2 \, \text{cm/sec}\) is: \[ \boxed{4 \, \text{cm}} \]