Question

With block C placed on top of B. the system undergoes simple harmonic motion with an amplitude of 0.12 m. Block B has a speed of 0.22 m/s at a displacement of 0.075 m from its equilibrium position. Determine the period of the motion. Express your answer with the appropriate units. What minimum value for the coefficient of static friction mu_s between B and C is needed if C is never to slip? Ignore any friction between B and the horizontal surface.

Answer 1

We know the amplitude, so, we know the total energy of the system:

$$E_T=\frac12 kA^2.$$

This is equivalent to the sum of elastic potential and kinetic energy of the system at the moment when the speed is 0.22 m/s and displacement of 0.075 m.

$$E_T=\frac12 kx^2+\frac12Mv^2.$$

The period of motion is

$$T=2\pi\sqrt{\frac{M}{k}}.$$

So, what we have is

$$\frac12 kA^2=\frac12 kx^2+\frac12Mv^2,\\\space\\ k=\frac{Mv^2}{A^2+x^2}.$$

Substitute this into the equation for period:

$$T=\frac{2\pi}{v}\sqrt{A^2+x^2}=4\text{ s}.$$

Now, let's talk about the coefficient of friction. Write the equation for displacement:

$$x(t)=A\sin\bigg(\frac{2\pi}{T}t\bigg),$$

the acceleration is the second derivative of displacement by time:

$$-a(t)=A\bigg(\frac{2\pi}{T}\bigg)^2\sin\bigg(\frac{2\pi}{T}t\bigg).$$

So, the magnitude of acceleration is

$$|a|=A\bigg(\frac{2\pi}{T}\bigg)^2=0.3\text{ m/s}^2.$$

The coefficient of friction is

$$\mu_s=\frac ag=0.03.$$