Single Choice

An oscillator of mass $$M$$ is at rest in its equilibrium position in a potential $$V = \dfrac {1}{2} k (x - X)^{2}$$. A particle of mass $$m$$ comes from right with speed $$u$$ and collides completely inelastically with $$M$$ and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after $$13$$ collisions is: $$(M = 10, m = 5, u = 1, k = 1)$$.

A$$\dfrac {1}{2}$$
B$$\dfrac {1}{\sqrt {3}}$$
Correct Answer
C$$\dfrac {2}{3}$$
D$$\sqrt {\dfrac {3}{5}}$$

Solution

Initial momentum of mass 'm' = mu =5

Final momentum of system= (M+m)v=mu = 5

For second collision, mass (m=5, u = 1) coming from right strikes with system of mass 15, both momentum have opposite direction.

$$\therefore$$ net momentum = zero

Similarly for $$12^{th}$$ collision momentum is zero.

For $$13^{th}$$ collision, total mass = 10 +12 $$\times $$ 5 = 70

Using conservation of momentum

$$70 \times 0 + 5 \times 1 = (70 + 5) v'$$

$$v' = \dfrac{1}{15}$$

Total mass $$=10+13\times 5=75$$

Finald KE of system $$=\displaystyle\frac{1}{2}mv^2=\frac{1}{2}\times 75\times \left[\displaystyle\frac{1}{15}\right]\left[\displaystyle\frac{1}{15}\right]$$

$$\displaystyle\frac{1}{2}k$$ $$A^2=\displaystyle\frac{1}{2}$$ $$\displaystyle 75\times \frac{1}{15}\times \frac{1}{15}$$

$$=\displaystyle\frac{1}{7}\times (1)A^2=\frac{1}{2}75\times \frac{1}{15}\times \frac{1}{15}$$

$$A^2=\frac{1}{3}$$

$$A=\displaystyle \frac{1}{\sqrt{3}}$$

SIMILAR QUESTIONS

Simple Harmonic Motion

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion 10 $$S^{-1}$$ . At, t = 0 the displacement is 5 m. What is the maximum acceleration? The initial phase is $$ \frac {\pi} {4}$$ .

Simple Harmonic Motion

In an engine the piston undergoes vertical simple harmonic motion with amplitude $$7\ cm$$. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :

Simple Harmonic Motion

A particle is executing a simple harmonic motion. Its maximum acceleration is $$\alpha$$ and maximum velocity is $$\beta$$. Then, its time period of vibration will be:

Simple Harmonic Motion

An object of mass '$$m$$' is kept on a horizontal platform. The platform is oscillating vertically with amplitude '$$a$$' and period '$$T$$'. The weight of object at lowest position is:

Simple Harmonic Motion

A $$5$$ kg collar is attached to a spring of spring constant $$500 Nm^{-1}$$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $$10$$cm and released. The maximum acceleration of the collar is

Simple Harmonic Motion

A particle executing simple harmonic motion with an amplitude A and angular frequency $$\omega$$. The ratio of maximum acceleration to the maximum velocity of the particle is

Simple Harmonic Motion

A particle executing SHM according to the equation $$x = 5 cos[2 \pi t + \frac{\pi}{4}]$$ in SI units. The displacement and acceleration of the particle at t = $$1.5$$s is:

Simple Harmonic Motion

A mass oscillates along the x-axis according to the law, $$x = x_0 cos (\omega t - \frac{\pi}{4})$$. If the acceleration of the particle is written as $$a = A cos (\omega t + \delta)$$, then

Simple Harmonic Motion

A particle is executing simple harmonic motion with amplitude of $$0.1 m$$. At a certain instant when its displacement is $$0.02$$ and its acceleration is $$0.5 { m }/{ { s }^{ 2 } }$$, the maximum velocity of the particle is (in $${ m }/{ s }$$) :

Simple Harmonic Motion

A particle is executing simple harmonic motion with an amplitude of 0.02 meter and frequency 50$$\mathrm { Hz }$$ The maximum acceleration of the particle is

Contact Details