Oscillations practice · Mixed difficulty

A uniform spring has an unstreched length $\lambda$ and a force constant $\mathrm{k}$. The spring is cut into two parts of unstreched length $\lambda_1$ and $\lambda_2$ such that $\lambda_1=\eta \lambda_2$ where $\eta$ is an integer. The corresponding force constants $\mathrm{k}_1$ and $\mathrm{k}_2$ are :

In the equation $F=k x$, the force constant $k$ does NOT depend upon:

In damped oscillations, the amplitude of oscillations is reduced to one-third of its initial value $\mathrm{a}_0$ at the end of 100 oscillations. Initial energy is $\mathrm{E}_0$. When the oscillator completes 200 oscillations, its energy must be

A body of mass $0.15 \mathrm{~kg}$ moving with a velocity of $15 \mathrm{~m}.\mathrm{s}^{-1}$ comes to rest when it hits a spring that is fixed at another end. If the force constant of the spring is $1500 \mathrm{~N}.\mathrm{m}^{-1}$, the compression in the spring is

At mean point of a body executing simple
harmonic motion the velocity will be

A mass $m$ is suspended from a spring. Its frequency of oscillation is $f$. The spring is cut into two halves and the same mass is suspended from one of the two pieces of the spring. The frequency of oscillation of the mass will be

A body of mass $10 \mathrm{~g}$ is executing simple harmonic motion about a point with an amplitude $20 \mathrm{~cm}$. If its maximum velocity is $100 \mathrm{~cm}.\mathrm{s}^{-1}$, its velocity will be $50 \mathrm{~cm} .\mathrm{s}^{-1}$ at a distance ____ $\mathrm{cm}$ from its mean position.

A particle is vibrating simple harmonically with an amplitude ' $a '$. The displacement of the particle when its energy is half kinetic and half potential is :

A particle is making simple harmonic motion along the $X$-axis. If at a distances ${x_1}$ and ${x_2}$ from the mean position, the velocities of the particle are ${v_1}$ and ${v_2}$ respectively, then the time period of its oscillation is given as

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

Assertion (A): The periodic time of hard spring is less a compared to that of soft spring.
Reason (R): The periodic time depends upon the spring constant.

A second's pendulum clock has a steel wire. The clock shows correct time at $25^{\circ} \mathrm{C}$. How much time does the clock loss or gain, in one week, when the temperature in increased to $35^{\circ} \mathrm{C}$ ?

$\left(\alpha_{\text {steel }}=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}\right)$

A particle executing SHM is given by equation $x=A \cos (\omega t+\phi)$. When particle is at half the amplitude with positive velocity initially, the initial phase $\phi$ is

Statement-I: Oscillations in spring mass system are due to continuous exchange of energy between the block and the spring.
Statement-II: The total energy of the spring mass system remains constant.

Assertion: Acceleration is proportional to the displacement. This condition is not sufficient for motion in simple harmonic.
Reason: In simple harmonic motion direction of displacement is also considered.

The function $\sin ^2(\omega t)$ represents:

$T_1\, T_2$ and $T_3$ are the time periods of a given pendulum on the surface of earth, at an altitude ' $h$ ' and at a depth ' $h$ ' below the earth's surface respectively then

A particle executes $SHM$ with an amplitude of $5\,cm$. The percentage of its energy which is potential energy at a displacement of $1\,cm$ from the mean position is

A particle performing S.H.M. starts from equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is $\pi \mathrm{~m} / \mathrm{s}$. Amplitude of oscillation is $\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)$

A body undergoes SHM according to the equation $x = (5m)\cos \left[ {2\pi t(s) + \frac{\pi }{4}} \right]$. The frequency of kinetic energy of body will be

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are ${{K_1}}$ and ${{K_2}}$ . The time period of the suspended mass will be

S-I: A particle executes SHM with frequency $f$. The frequency with which total mechanical energy of particle oscillates is $2 f$.
S-II: Both kinetic as well as potential energy of particle in SHM oscillate with a frequency $2 f.$

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

A particle moves so that its acceleration a is given by $a = - bx$ where $x$ is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is

When an oscillator completes 50 oscillations, its amplitude reduces to $\frac{1}{3}$ times of the initial value. The amplitude, when it completes 100 oscillations will be

Under the action of a force $F=-k x^3$, the motion of a particle is $(k=$ a positive constant $)$

A solid sphere of radius $R$ and density $\rho$ is attached to one end of a massless spring of force constant $k$. The other end of the spring is connected to another solid sphere of radius $R$ and density $3 \rho$. The complete arrangement is placed in a liquid of density $2 \rho$ and is allowed to reach equilibrium. The correct statement is

Two particles are executing S.H.M. The equations of their motion are $y_1=10 \sin \left(\omega t+\frac{\pi T}{4}\right), y_2=25 \sin \left(\omega t+\frac{\sqrt{3} \pi T}{4}\right)$.
What is the ratio of their amplitude?

When the displacement is half the amplitude, the ratio of potential energy to the total energy is

The displacement y of a particle executing periodic motion is given by $y = 4{\cos ^2}\left( {t/2} \right)\sin \left( {1000t} \right)$ . This expression may be considered to be a result of the superposition of…….. independent harmonic motions