Oscillations practice · Mixed difficulty

A mass $M$ is suspended from a massless spring. An additional mass $m$ stretches the spring further by a distance $x$. The combined mass will oscillate with a period

A body is moving in a room with a velocity of 20 m/s perpendicular to the two walls separated by 5 meters. There is no friction and the collisions with the walls are elastic. The motion of the body is

The displacement of a particle varies according to the relation $x=3 \sin 100 t+8 \cos ^2 50 t$. Which of the following is incorrect about this motion.

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

A particle is executing SHM along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $v_1$ and $v_2$, respectively. Its time period is

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force. $F = {F_0}\sin \omega t$ If the amplitude of the particle is maximum for ω=${\omega _1}$ and the energy of the particle is maximum for ω=${\omega _2}$ then

In a damped oscillation with damping constant $b,$
the time taken for its mechanical energy to drop
to half its value is

The amplitude of vibration of a particle is given by ${a_m} = ({a_0})/(a{\omega ^2} - b\omega + c);$ where ${a_0},a,b$ and c are positive. The condition for a single resonant frequency is

A particle is performing SHM with amplitude '$A$'. Find out the position where velocity is half of the maximum.

A particle is executing S.H.M. Then the graph of acceleration as a function of displacement is

In a simple pendulum, the period of oscillation $T$ is related to length of the pendulum $l$ as

Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock 2 is placed on a space station located at a height $h$ above the earth surface. Clock $-1$ and Clock - 2 operate at time periods $4 \mathrm{~s}$ and $6 \mathrm{~s}$ respectively. Then the value of $h$ is-

(consider radius of earth $R_E=6400 \mathrm{~km}$ and $g$ on earth $10 \mathrm{~m} / \mathrm{s}^2$ )

Two particles are executing SHMs. The equations of their motions are

$y_1=10 \sin \left(\omega t+\frac{\pi}{4}\right) \text { and } y_2=5 \sin \left(\omega t+\frac{\sqrt{3} \pi}{4}\right)$

What is the ratio of their amplitudes?

If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

The potential energy of a particle (${U_x}$) executing SHM is given by

The bob of a simple pendulum of length L is released at time $t\mathrm{ } = \mathrm{ }0$ from a position of small angular displacement. Its linear displacement at time t is given by

The length of a seconds pendulum at a height $h=2 R$ from earth surface will be
(Given $R=$ Radius of earth and acceleration due to gravity at the surface of earth, $g=\pi^2 \mathrm{~ms}^{-2}$ )

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10{s^{ - 1}}$. At, t= 0 the displacement is 5m. What is the maximum acceleration? (The initial phase is $\frac{\pi }{4}$)

A particle is subjected simultaneously to two SHM’s one along the $x$-axis and the other along the $y$ -axis. The two vibrations are in phase and have unequal amplitudes. The particle will execute

Statement-1: For a particle of mass $1 \mathrm{~kg}$, executing S.H.M., if slope of restoring force vs displacement graph is $=-1$, then the time period of oscillation will be $6.28 \mathrm{~s}$.
Statement-2: If $1 \mathrm{~kg}$ mass is replaced by $2 \mathrm{~kg}$ mass and rest of the information remains same as in statement-1, then the time period of oscillation will remain $6.28 \mathrm{~s}$.

A light spring of force constant $8 \mathrm{Nm}^{-1}$ is cut into two equal halves and the two are connected in parallel; the equivalent force constant of the system is

If a simple pendulum executing SHM falls freely along with the support, then

A particle executing simple harmonic motion along $Y$-axis has its motion described by the equation $y=A \sin (\omega t)+B$. The amplitude of the simple harmonic motion is

A particle is subjected to two simple harmonic motions along ${x}$ and ${y}$ directions according to $x=3 \sin 100 \pi t$ and $y=4 \sin 100 \pi t$

S-I: If the amplitude of the oscillation is doubled the maximum kinetic energy becomes four times.
S-II: If the amplitude of the oscillation is doubled the minimum potential energy also becomes four times.

A mass of $0.2 \mathrm{~kg}$ is attached to the lower end of a massless spring of force constant $200 \mathrm{Nm}^{-1}$, the upper end of which is fixed to a rigid support. Which of the following statement(s) is/are true?

The motion of particle executing S.H.M is given by $x = 0.01\,\,\sin \pi (100\,t + 0.05),$ where x in metres and time is in seconds. The time period is

The vertical motion of a ship at sea is described by the equation $\frac{d^2 x}{d t^2}=-4 x$, where $x$ is the vertical height of the ship (in metre) above its mean position. If it oscillates through a height of $1 \mathrm{~m}$, its maximum vertical

If the length of second's pendulum is decreased by $2 \%$, how many seconds, it will lose per day?

The potential energy of a particle with displacement X is U(X). The motion is simple harmonic, when (K is a positive constant)