Question

The force (F in newton) acting on a particle of mass 90 g executing simple harmonic motion is given by $F+0.04\pi^2y=0$, where y is displacement of the particle in meter. If the amplitude of the particle is $\frac{6}{\pi}$ m, then the maximum velocity of the particle is

• A. 6 ms$^{-1}$
• B. 2 ms$^{-1}$
• C. 8 ms$^{-1}$
• D. 4 ms$^{-1}$

Hint:

Any motion described by the equation $F = -(\text{positive constant}) \times y$ is Simple Harmonic Motion. The constant of proportionality is the spring constant $k$. From there, you can find the angular frequency $\omega = \sqrt{k/m}$ and other SHM parameters.

Answer 1

The Correct Option is D

Step 1: Relate the given force equation to the standard SHM force equation.
The equation for a particle in Simple Harmonic Motion (SHM) is $F = -ky$, where $k$ is the force constant. The given equation is $F+0.04\pi^2y=0$, which can be rewritten as: \[ F = -(0.04\pi^2)y. \] By comparing the two equations, we find the force constant: $k = 0.04\pi^2$ N/m.

Step 2: Calculate the angular frequency $\omega$.
The angular frequency $\omega$ is related to the force constant $k$ and mass $m$ by the formula $\omega = \sqrt{k/m}$. The mass is given as $m = 90$ g $= 0.09$ kg. \[ \omega = \sqrt{\frac{0.04\pi^2}{0.09}} = \sqrt{\frac{4\pi^2}{9}} = \frac{2\pi}{3} \text{ rad/s}. \]

Step 3: State the formula for maximum velocity in SHM.
The maximum velocity ($v_{max}$) of a particle in SHM is given by the product of its amplitude ($A$) and its angular frequency ($\omega$). \[ v_{max} = A\omega. \]

Step 4: Calculate the maximum velocity.
We are given the amplitude $A = \frac{6}{\pi}$ m. \[ v_{max} = \left(\frac{6}{\pi}\right) \left(\frac{2\pi}{3}\right). \] The $\pi$ terms cancel out. \[ v_{max} = \frac{6 \times 2}{3} = \frac{12}{3} = 4 \text{ m/s}. \] \[ \boxed{v_{max} = 4 \text{ m/s}}. \]