A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $\omega$. If the radius of the bottle is 2.5 cm then $\omega$ is close to: (density of water $= 10 ^ { 3 } \mathrm {~kg} / \mathrm { m } ^ { 3 }$)
(1) $5.00 \mathrm { rad } \mathrm { sec } ^ { - 1 }$
(2) $2.50 \mathrm { rad } \mathrm { sec } ^ { - 1 }$
(3) $7.9 \mathrm { rad } \mathrm { sec } ^ { - 1 }$
(4) $3.75 \mathrm { rad } \mathrm { sec } ^ { - 1 }$

A simple pendulum of length 1 m is oscillating with an angular frequency $10 \mathrm { rad } / \mathrm { s }$. The support of the pendulum starts oscillating up and down with a small angular frequency of $1 \mathrm { rad } / \mathrm { s }$ and an amplitude of $10 ^ { - 2 } \mathrm {~m}$. The relative change in the angular frequency of the pendulum is best given by:
(1) $10 ^ { - 3 } \mathrm { rad } / \mathrm { s }$
(2) $1 \mathrm { rad } / \mathrm { s }$
(3) $10 ^ { - 1 } \mathrm { rad } / \mathrm { s }$
(4) $10 ^ { - 5 } \mathrm { rad } / \mathrm { s }$

As shown in the figure, a bob of mass $m$ is tied to a massless string whose other end portion is wound on a fly wheel (disc) of radius $r$ and mass $m$. When released from rest the bob starts falling vertically. When it has covered a distance of $h$, the angular speed of the wheel will be:
(1) $\frac { 1 } { \mathrm { r } } \sqrt { \frac { 4 \mathrm { gh } } { 3 } }$
(2) $r \sqrt { \frac { 3 } { 2 g h } }$
(3) $\frac { 1 } { \mathrm { r } } \sqrt { \frac { 2 \mathrm { gh } } { 3 } }$
(4) $r \sqrt { \frac { 3 } { 4 g h } }$

A particle starts executing simple harmonic motion (SHM) of amplitude $a$ and total energy $E$. At any instant, its kinetic energy is $\frac { 3 E } { 4 }$, then its displacement $y$ is given by:
(1) $y = a$
(2) $y = \frac { a } { \sqrt { 2 } }$
(3) $y = \frac { a \sqrt { 3 } } { 2 }$
(4) $y = \frac { a } { 2 }$

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?
(1) $x = 0$
(2) $x = \pm A$
(3) $x = \pm \frac { A } { \sqrt { 2 } }$
(4) $x = \frac { A } { 2 }$

An object of mass 0.5 kg is executing simple harmonic motion. Its amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $\mathrm { t } = \frac { \mathrm { T } } { 4 } \mathrm {~s}$ starting from mean position. Assume that the initial phase of the oscillation is zero.
(1) 0.62 J
(2) $6.2 \times 10 ^ { - 3 } \mathrm {~J}$
(3) $1.2 \times 10 ^ { 3 } \mathrm {~J}$
(4) $6.2 \times 10 ^ { 3 } \mathrm {~J}$

In the reported figure, two bodies $A$ and $B$ of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be \_\_\_\_ rad $\mathrm{s}^{-1}$ when $k = 20\mathrm{~N~m}^{-1}$.

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is $\frac { T _ { b } } { T _ { a } } = \sqrt { x }$, where value of $x$ is $\_\_\_\_$. (Round off to the Nearest Integer)

A particle of mass 1 mg and charge $q$ is lying at the mid-point of two stationary particles kept at a distance 2 m when each is carrying same charge $q$. If the free charged particle is displaced from its equilibrium position through distance $x$ ($x \ll 1\mathrm{~m}$), the particle executes SHM. Its angular frequency of oscillation will be \_\_\_\_ $\times 10^{5}\mathrm{~rad~s}^{-1}$ (if $q^{2} = 10\mathrm{~C}^{2}$).

A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac { 1 } { a } \mathrm {~s}$. The value of $a$ to the nearest integer is

A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is $\sqrt { x } \mathrm {~cm}$. The value of $x$ is $\_\_\_\_$ .

As per given figures, two springs of spring constants $K$ and $2K$ are connected to mass $m$. If the period of oscillation in figure (a) is 3 s, then the period of oscillation in figure (b) will be $\sqrt { x }$ s. The value of $x$ is $\_\_\_\_$.

A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K_1$ and $K_2$. For the frictionless surface, the time period of oscillation of mass $m$ is
(1) $2\pi\sqrt{\frac{m}{K_1 + K_2}}$
(2) $\frac{1}{2\pi}\sqrt{\frac{K_1 - K_2}{m}}$
(3) $2\pi\sqrt{\frac{m}{K_1 - K_2}}$
(4) $\frac{1}{2\pi}\sqrt{\frac{K_1 + K_2}{m}}$

A circular plate is rotating in horizontal plane, about an axis passing through its centre and perpendicular to the plate, with an angular velocity $\omega$. A person sits at the centre having two dumbbells in his hands. When he stretched out his hands, the moment of inertia of the system becomes triple. If $E$ be the initial Kinetic energy of the system, then final Kinetic energy will be $\frac { E } { x }$. The value of $x$ is

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement ($x$) starting from mean position to extreme position ($A$) is given by
(1) [Figure]
(2) [Figure]
(3) [Figure]
(4) [Figure]

A particle executes SHM of amplitude $A$. The distance from the mean position when its kinetic energy becomes equal to its potential energy is:
(1) $\frac { 1 } { \sqrt { 2 } } A$
(2) $2 A$
(3) $\sqrt { 2 A }$
(4) $\frac { 1 } { 2 } A$

The general displacement of a simple harmonic oscillator is $x = A \sin \omega t$. Let $T$ be its time period. The slope of its potential energy $(U)$ - time $(t)$ curve will be maximum when $t = \frac{T}{\beta}$. The value of $\beta$ is $\_\_\_\_$.

A rectangular block of mass 5 kg attached to a horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14 s . The maximum force exerted by spring on block is $\_\_\_\_$ N.

Consider a disc of mass 5 kg , radius 2 m , rotating with angular velocity of $10 \mathrm { rad } \mathrm { s } ^ { - 1 }$ about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is $\_\_\_\_$ J.

A simple harmonic oscillator has an amplitude $A$ and time period $6 \pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x = A$ to $x = \frac { \sqrt { 3 } } { 2 } A$ will be $\frac { \pi } { x } \mathrm {~s}$, where $x =$ $\_\_\_\_$ .

The time period of simple harmonic motion of mass $M$ in the given figure is $\pi\sqrt{\dfrac{\alpha M}{5K}}$, where the value of $\alpha$ is $\_\_\_\_$.

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4 \mathrm {~m} , 2 \mathrm {~ms} ^ { - 1 }$ and $16 \mathrm {~ms} ^ { - 2 }$ at a certain instant. The amplitude of the motion is $\sqrt { x } , \mathrm {~m}$ where $x$ is $\_\_\_\_$

Q24. The displacement of a particle executing SHM is given by $x = 10 \sin \left( w t + \frac { \pi } { 3 } \right) m$. The time period of motion is 3.14 s . The velocity of the particle at $t = 0$ is $\_\_\_\_$ $\mathrm { m } / \mathrm { s }$.