Oscillations Test

Directions For Questions

A $$2\ kg$$ block hangs without vibrating at the bottom end of a spring with a force constant of $$800\ N/m$$. The top end of the spring is attached to the ceiling of an elevator car. The car is rising with an upwards acceleration of $$10\ m/s^{2}$$ when the acceleration suddenly ceases at time $$t=0$$ and the car moves upward with constant speed $$(g=10\ m/s^{2})$$

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One end of an ideal spring is fixed to a wall at origin $$O$$ and the axis of spring is parallel to the $$x-$$ axis. A block of mass $$m=1\ kg$$ is attached to free end of the spring and it is performing $$SHM$$. Equation fo position of the block in coordinates system shown in Fig $$4.145$$ is $$x=10+3\sin (10\ t)$$, where $$t$$ is in second and $$x$$ in $$cm$$.
Another block of mass $$M=3\ kg$$, moving towards the origin with velocity $$30\ cm/s$$ collides with the block performing $$SHM$$ at $$t=0$$ and gets stuck to it.

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A block of mass $$M$$ is suspended from on end of a light spring as shown. The origin $$O$$ is considered at distance equal to the natural length of the spring from the ceiling and vertical downward direction as a positive $$y-$$ axis. When the system is in equilibrium, a bullet of mass $$m/3$$ moving in a vertically upward direction with velocity $$v_{0}$$ strikes the block and embeds into it. As a result, the block (with a bullet embedded into it) moves up and starts oscillating.
Based on the given information answer the following question:

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A block of mass $$m$$ is connected to a spring of spring constant $$k$$ as shown in Fig $$4.147$$. The block is found at its equilibrium position at $$t=1\ s$$ and it has a velocity of $$+0.25\ m/s$$ at $$t=2\ s$$. The time period oscillation is $$6\ s$$.
Based on the given information answer the following question:

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In a physical pendulum the time period for small oscillation is given by, $$T=2\pi\sqrt{I/Mgd}$$ where $$I$$ is the moment of inertia of the body about an axis passing through a pivoted point $$O$$ and perpendicular to the plane of oscillation and $$d$$ is the separation point between center of gravity and the pivoted point.
In the physical pendulum, a special point exists where if we concentrate the entire mass of the body the resulting simple pendulum (w.r.t. pivot point $$O$$) will have the same time period as that of the physical pendulum. This point is termed the center of oscillation.
$$T=2\pi\sqrt{\dfrac{I}{Mgd}}=2\pi\sqrt{\dfrac{L}{g}}$$
Moreover, this point possesses two other important remarkable properties:
Property I: Time period of the physical pendulum about the center of oscillation (if it would be pivoted) is the same as in the original case.
Property II: If an impulse is applied at the center of oscillation in the plane of oscillation the effect of this impulse at a pivoted point is zero. Because of this property, this point is also known as the center of percussion.
From the given information answer the following questions:

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