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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.

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Solution

Let x1 and x2 be the amplitudes of oscillation of masses m and M respectively.

(a) As the centre of mass should not change during the motion, we can write:
mx1 = Mx2 …(1)

Let k be the spring constant. By conservation of energy, we have:
12kx02=12kx1+x22 …(2)
where x0 is the length to which spring is stretched.

From equation (2) we have,
x0=x1+x2

On substituting the value of x2 from equation (1) in equation (2), we get:
x0=x1+mx1M⇒x0=1+mMx1⇒x1=MM+mx0

Now, x2=x0-x1
On substituting the value of x1 from above equation, we get:
⇒ x2=x01-MM+m⇒ x2=mx0M+m

Thus, the amplitude of the simple harmonic motion of a car, as seen from the road is mx0M+m.


(b) At any position,
Let v1 and v2 be the velocities.

Using law of conservation of energy we have,
12Mv2+12mv1-v22+12kx1+x22=constant ...3

Here, (v1 − v2) is the absolute velocity of mass m as seen from the road.

Now, from the principle of conservation of momentum, we have:
Mx2 = mx1

⇒x1=Mmx2 ....4Mv2=mv1-v2⇒v1-v2=Mmv2 ....5

Putting the above values in equation (3), we get:
12Mv22+12mM2m2v22+12kx221+Mm2=constant∴M1+Mmv22+k1+Mmx22=constant⇒Mv22+k1+Mmx22=constant

Taking derivative of both the sides, we get:
M×2v2dv2dt+kM+mm2x2dx2dt=0⇒ma2+kM+mmx2=0 because, v2=dx2dta2x2=-kM+mMm=ω2∴ω=kM+mMmTherefore, time period, T=2πMmkM+m


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