Question

A horizontal board is performing simple harmonic oscillations horizontally with an amplitude of \(0.3 \, {m}\) and a period of \(4 \, {s}\). The minimum coefficient of friction between a heavy body placed on the board if the body is not to slip is:

• A. \( \mu = 0.05 \)
• B. \( \mu = 0.075 \)
• C. \( \mu = 0.173 \)
• D. \( \mu = 1.14 \)

Hint:

When dealing with simple harmonic motion and friction, always consider the maximum acceleration and ensure that the frictional force is sufficient to counteract this maximum force.

Answer 1

The Correct Option is B

The minimum coefficient of friction \( \mu \) required to prevent slipping is determined by the maximum acceleration \( a_{{max}} \) the body experiences during the oscillations. For simple harmonic motion (SHM), the maximum acceleration is given by: \[ a_{{max}} = \omega^2 A \] where \( \omega \) is the angular frequency and \( A \) is the amplitude. The angular frequency \( \omega \) can be calculated from the period \( T \) as: \[ \omega = \frac{2\pi}{T} \] Substituting \( T = 4 \, {s} \) and \( A = 0.3 \, {m} \), we find: \[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \, {rad/s} \] \[ a_{{max}} = \left(\frac{\pi}{2}\right)^2 \times 0.3 = \frac{\pi^2}{4} \times 0.3 \] The frictional force needed to prevent slipping is \( f = \mu mg \), equating this to the necessary centripetal force \( ma_{{max}} \), we have: \[ \mu = \frac{a_{{max}}}{g} \] Given \( g = 9.81 \, {m/s}^2 \), substituting the values: \[ \mu = \frac{\frac{\pi^2}{4} \times 0.3}{9.81} \] \[ \mu \approx 0.075 \]