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Oscillations Test - 80

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Oscillations Test - 80
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  • Question 1
    1 / -0
    Between the plates of the capacitor with potential difference $$V$$ across its plate such that upper plate is $$-ve$$, a ball with positive charge '$$q$$' and mass '$$m$$' is suspended by a thread of length '$$l$$'. If the electrostatic force acting on a ball is less than the gravitational force, what should be the period of the ball?

    Solution

  • Question 2
    1 / -0
    A particle of mass m moves in  a one-dimensional potential energy $$ U(x) = -ax^2 +bx^4 $$, where a and 'b' are positive constants. the angular frequency of small oscillation about the minima of the potential energy is equal to
    Solution

  • Question 3
    1 / -0
    A block of mass $$m$$ is pushed against a spring whose spring constant is $$k$$ fixed at one end with a wall. The block can slide on a frictionless table as shown in figure. If the natural length of spring is $$L_0$$ and it is compressed to half its length when the block is released, find the velocity of the block, when the spring has natural length.

    Solution

  • Question 4
    1 / -0

    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 $$(g=10\ m/s^{2})$$

    ...view full instructions

    What is the angular frequency of oscillation of the block after the acceleration ceases?
    Solution
    As  elevator starts rising up with $$a=10\,m/s^2$$ pseudo force act on block.
    $$F=ma$$ in downward direction.
    Spring Force =$$Kx$$ in upward direction
    angular frequency is given by
    $$\omega=\sqrt{\dfrac{k}{m}}$$
    $$\omega=\sqrt{\dfrac{800}{2}}$$
    $$\omega=20\,rad/s$$
  • Question 5
    1 / -0

    Directions For Questions

    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.

    ...view full instructions

    New amplitude of oscillation is
    Solution
    Conserving linear momentum.
    $$(1+3)v=1\times 0.3+3(-0.3)$$
    $$v=-0.15\ m/s$$
    Negative sign indicates that combined body starts to move leftward. but at the instant of collision, spring is in its natural length or combined body is in equilibrium position. Hence at $$t=0$$, phase of combined body becomes equal to $$\pi$$.
    $$\therefore$$ New amplitude of oscillation is
    $$a'=\dfrac{|v|}{\omega}=\dfrac{0.15}{5}=0.03\ m=3\ m$$
  • Question 6
    1 / -0

    Directions For Questions

    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:

    ...view full instructions

    Mark out the correct statement(s).
    Solution

  • Question 7
    1 / -0
    A particle performs simple harmonic motion with amplitude $$A$$ and time period $$T$$. The mean velocity of the particle over the time in interval which it travels a distance of $$A/2$$ starting from executing position is 
    Solution

  • Question 8
    1 / -0

    Directions For Questions

    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:

    ...view full instructions

    Determine the velocity of particle at $$t=5\ s$$
    Solution

  • Question 9
    1 / -0
    A spring balance has a scale that can read from $$0$$ to $$50\ kg$$ . The length of the scale is $$20\ cm$$. A body suspended from this balance when displaced and released oscillates harmonically with a time period of $$0.6\ s$$. The mass of the body is ( take $$g=10\ m/s^2)$$
    Solution

  • Question 10
    1 / -0

    Directions For Questions

    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:

    ...view full instructions

    For the above question locate the centre of oscillation.
    Solution

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