A particle of mass is executing oscillations about the origin on the x-axis. Its potential energy is V(x) =k | x |³, where k is a positive constant. If the amplitude of oscillation is a,then its time period T is

proportional to ¹⁄_√2
proportional to √a
independent a^³⁄₂
None of these

Solution

A uniform pole of length l = 2 L is laid on smooth horizontal table as shown in figure. The mass of pole is M and it is connected to a friction less axis at O.A spring with force constant k is connected to the other end. The pole is displaced by a small angle θ₀ from equilibrium position and released such that it performs small oscillations. Then

\(\omega _{0}=\sqrt{\frac{M}{3k}}\)
\(\omega _{0}=\sqrt{\frac{k}{3m}}\)
\(\omega _{0}=\sqrt{\frac{3k}{M}}\)
\(\omega _{0}=\sqrt{\frac{k}{2M}}\)

Solution

On Earth, a body suspended on a spring of negligible mass causes extension L and undergoes oscillations along length of the spring with frequency f. On the Moon, the same quantities are L/n and f ‘ respectively. The ratio f ‘/f is

n
¹⁄_n
n^–1/2
1

Solution

Oscillations along spring length are independent of gravitation.

A block rests on a horizontal table which is executing SHM in the horizontal plane with an amplitude ‘a’. If the coefficient of friction is ‘μ’, then the block just starts to slip when the frequency of oscillation is

\(\frac{1}{2\pi }\sqrt{\frac{\mu g}{a}}\)
\(\sqrt{\frac{\mu g}{a}}\)
\(2\pi \sqrt{\frac{a}{\mu g}}\)
\(\sqrt{\frac{a}{\mu g}}\)

Solution

The time period of a simple pendulum of infinite length is(Re= radius of Earth)

\(T=2\pi \sqrt{\frac{R_{e}}{g}}\)
\(T=2\pi \sqrt{\frac{2R_{e}}{g}}\)
\(T=2\pi \sqrt{\frac{R_{e}}{2g}}\)
T = ∞

Solution

The graph shown in figure represents

S.H.M.
circular motion
rectillinear motion
uniform circular motion

Solution

t = 0, v maximum. The motion begins from mean position. So it represents S.H.M.

A mass m fall on spring of spring constant k and negligible mass from a height h. Assuming it sticks to the pan and executes simple harmonic motion, the maximum height upto which the pan will rise is

^mg⁄_k
\(\frac{mg}{k}\left [ \sqrt{1+\frac{2kh}{mg}-1} \right ]\)
\(\frac{mg}{k}\left [ \sqrt{1+\frac{2kh}{mg}+1} \right ]\)
\(\frac{mg}{k}\left [ \sqrt{1+\frac{kh}{mg}-1} \right ]\)

Solution

A pendulum bob is raised to a height h and released from rest. At what height will it attain half of its maximum speed?

^3h⁄₄
^h⁄₂
^h⁄₄
0.707h

Solution

Frequency of oscillation is proportional to

\(\sqrt{\frac{3k}{m}}\)
\(\sqrt{\frac{k}{m}}\)
\(\sqrt{\frac{2k}{m}}\)
\(\sqrt{\frac{m}{3k}}\)

Solution

A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l. The system is rotated about the other end of the spring with an angular velocity ω, in gravity free space. The increase in length of the spring will be

\(\frac{m\omega ^{2}\iota }{k}\)
\(\frac{m\omega ^{2}\iota }{k-m\omega ^{2}}\)
\(\frac{m\omega ^{2}\iota }{k+m\omega ^{2}}\)
None of these

Solution

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