Question

All the surfaces shown in figure (12−E15) are frictionless. The mass of the care is M, that of the block is m and the spring has spring constant k. Initially, the car and the block are at rest and the spring is stretched through x length x0 when the system is released. (a) Find the amplitudes of the simple harmonic motion of the block and of the care as seen from the road. (b) Find the time period(s) of the two simple harmonic motions.
Figure

Answer 1

Answer:

Amplitude of the two blocks will be

$x_1 = \frac{M}{m + M}x_o$

$x_2 = \frac{m}{m + M}x_o$

Part b)

Time period of SHM is

$T = 2\pi\sqrt{\frac{mM}{k(m + M)}}$

Explanation:

Let the amplitudes of two blocks are x1 and x2

so here we have

$x_o = x_1 + x_2$

also we know that

$mx_1 = Mx_2$]

so we will have

$x_1 = \frac{M}{m + M}x_o$

$x_2 = \frac{m}{m + M}x_o$

Part b)

Now for time period of the motion

$F = - kx$

$m\frac{d^2x_1}{dt^2} = - kx$

$M\frac{d^2x_2}{dt^2} = - kx$

so we have

$\frac{d^2x_1}{dt^2} + \frac{d^2x_2}{dt^2} = -(\frac{k}{m} + \frac{k}{M}) x$

so we will have

$\frac{d^2x}{dt^2} = -(\frac{k(m + M)}{mM})x$

so time period will be

$T = 2\pi\sqrt{\frac{mM}{k(m + M)}}$

#Learn

Topic : SHM

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