Question

A particle in SHM has a speed of \(6\,cm/s\) at the mean position and an amplitude of \(4\,cm\). Find its position when its velocity is \(2\,cm/s\).

• A. \( \frac{8\sqrt{2}}{3} \,cm \)
• B. \( \frac{4\sqrt{2}}{3} \,cm \)
• C. \( \frac{8}{3} \,cm \)
• D. \( 2\sqrt{2} \,cm \)

Hint:

Total mechanical energy in SHM stays constant, split between kinetic and potential energy at every position. Try writing this conservation equation between the mean position, where speed is maximum, and the point where the given velocity occurs — it lets you find the position without separately calculating the angular frequency.

Answer 1

The Correct Option is A

Concept: For a particle performing Simple Harmonic Motion (SHM), the velocity at any position \(x\) is given by \[ v = \omega \sqrt{A^2 - x^2} \] where
• \(A\) = amplitude
• \(x\) = displacement from mean position
• \(\omega\) = angular frequency At the mean position, velocity is maximum: \[ v_{max} = \omega A \]

Step 1: Find angular frequency. Given \[ v_{max} = 6\,cm/s \] \[ A = 4\,cm \] Using \[ v_{max} = \omega A \] \[ 6 = \omega \times 4 \] \[ \omega = \frac{3}{2} \]

Step 2: Use the SHM velocity formula. \[ v = \omega \sqrt{A^2 - x^2} \] Substitute \(v=2\): \[ 2 = \frac{3}{2}\sqrt{16 - x^2} \]

Step 3: Solve the equation. \[ \sqrt{16-x^2} = \frac{4}{3} \] Square both sides: \[ 16 - x^2 = \frac{16}{9} \] \[ x^2 = \frac{128}{9} \] \[ x = \frac{8\sqrt2}{3} \] Thus, the position of the particle is \[ \boxed{\frac{8\sqrt2}{3}\,cm} \]

Answer 2

Concept:

• In SHM, total mechanical energy stays constant at every instant: $\text{KE} + \text{PE} = \tfrac{1}{2}m\omega^2A^2$, since energy only converts between kinetic and potential forms.
• At the mean position, displacement is zero, so potential energy is zero and the entire energy is kinetic — this fixes the total energy in terms of the maximum speed.
• Writing this balance at a general displacement $x$ cancels the mass and directly relates $v$, $v_{max}$, $x$ and $A$, without separately computing the angular frequency $\omega$.

Step 1: Set up the energy balance between the mean position and displacement $x$.
$\tfrac{1}{2}mv_{max}^2 = \tfrac{1}{2}mv^2 + \tfrac{1}{2}m\omega^2x^2$
Cancelling $\tfrac{1}{2}m$: $v_{max}^2 = v^2 + \omega^2x^2$  ...(i)

Step 2: Eliminate $\omega$ using $v_{max} = \omega A$, so $\omega = v_{max}/A$. Substitute into (i):
$v_{max}^2 = v^2 + \dfrac{v_{max}^2}{A^2}x^2$
$v_{max}^2 - v^2 = \dfrac{v_{max}^2}{A^2}x^2 \Rightarrow x = \dfrac{A\sqrt{v_{max}^2 - v^2}}{v_{max}}$.

Step 3: Substitute the given values $v_{max} = 6\,cm/s$, $v = 2\,cm/s$, $A = 4\,cm$.
$x = \dfrac{4\sqrt{36-4}}{6} = \dfrac{4\sqrt{32}}{6} = \dfrac{4 \times 4\sqrt{2}}{6} = \dfrac{16\sqrt{2}}{6} = \dfrac{8\sqrt{2}}{3}$.

Final Answer: $x = \dfrac{8\sqrt{2}}{3}\,cm$.