Oscillations practice · Mixed difficulty

A particle executes simple harmonic motion between $x$ $= - A\,\,\mathrm{and}\,\,x = + A$ The time taken for it to go from $0\mathrm{ }to\;A/2\,\,is\,\,{T_1}$ and to go from $A/2$ to $A\,\,\mathrm{is}\,\,{T_2}$ Then

A particle undergoes SHM with a time period of 2 seconds. In how much time will it travel from its mean position to a displacement equal to half of its amplitude?

In a seconds pendulum, the mass of the bob is $30 \mathrm{~g}$. If it is replaced by $90 \mathrm{~g}$ mass, then its time period will be

The potential energy of a particle of mass $2 \mathrm{~kg}$ in SHM is $\left(9 x^2\right)$ joule. Here, $x$ is the displacement from mean position. If total mechanical energy of the particle is $36 \mathrm{~J}$, then maximum speed of the particle is

Maximum kinetic energy of a particle of mass $1 \mathrm{~kg}$ in SHM is $8 \mathrm{~J}$. Time period of SHM is $4 \mathrm{~s}$. Maximum potential energy during the motion is $10 \mathrm{~J}$. Then,

The equation of a damped simple harmonic motion is $m \frac{d^2 x}{d t^2}+b \frac{d x}{d t}+k x=0$. Then, the angular frequency of oscillation is

The amplitude and the periodic time of a SHM are $5 \mathrm{~cm}$ and $6 \mathrm{~s}$, respectively. At a distance of $2.5 \mathrm{~cm}$ away from the mean position, the phase will be

A conical pendulum of length 1 m makes an angle $\theta = {45^0}$ w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m. and its centre is vertically below O. The speed of the pendulum, in its circular path, will be $\left( {g = 10\mathrm{ }m{s^{ - 2}}} \right)$

The Q factor for a damped oscillator is given
by the

A spring has a natural length of $50 \mathrm{~cm}$ and a force constant of $2 \times 10^3 \mathrm{Nm}^{-1}$. A body of mass $10 \mathrm{~kg}$ is suspended from it and the spring is stretched. If the body is pulled down to a length of $58 \mathrm{~cm}$ and released, it executes simple harmonic motion. What is the net force on the body when it is at its lowermost position of its oscillation?

A horizontally placed spring mass system has time period $T$. The same system is now placed on a car moving with acceleration $a$ in horizontal direction. Then,

Assertion(A): In SHM, $v-x$ graph is an ellipse, where $v$ is velocity and $x$ is displacement from mean position.

Reason(R): Equation between $v$ and $x$ is $\frac{v^2}{\omega^2}+\frac{x^2}{1}=A^2$.

Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with period 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is

Two mutually perpendicular simple harmonic vibrations have same amplitude, frequency and phase. When they superimpose, the resultant from of vibration will be

The minimum phase difference between two simple harmonic oscillations,
${y_1} = \frac{1}{2}\sin \,\,\omega t + \frac{{\sqrt 3 }}{2}\cos \,\,\omega t$
${y_2} = \sin \,\,\omega t + \cos \,\,\omega t,$ is

An object of mass m is attached to light string which passes through a hollow tube. The object is set into rotation in a horizontal circle of radius , ${r_1}$ .If the string is pulled shortening the radius to ${r_2}$ ,the ratio of new kinetic energy to the original kinetic energy is

A linear harmonic oscillator of force constant $2 \times 10^6 \mathrm{Nm}^{-1}$ and amplitude $0.01 \mathrm{~m}$ has a total mechanical energy of $160 \mathrm{~J}$. Its

The value of phase at maximum displacement from the mean position of a particle in S.H.M. is:

The displacement $x$ of a particle in motion is given in terms of time by $x(x-4)=1-5 \cos \omega t$.

The maximum acceleration of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible when

Assertion(A): If a pendulum is suspended in a lift and lift accelerates upwards, then its time period will decrease.

Reason(R): Effective value of $g$ will be $g_{\text {eff }}=g+a$

A hollow sphere is filled with water through the small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flow out of the hole at the bottom, the period of oscillation will

A particle executing a simple harmonic motion has a period of 6 s. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

Starting from $y = A$ $\sin \,\,\omega t\mathrm{ }or\mathrm{ }y = A\,\,\cos \,\,\omega t$

Two SHM are represented by equations,
${y_1} = 6\cos \left( {6\pi t + \frac{\pi }{6}} \right),{y_2} = 3\left( {\sqrt 3 \sin 3\pi t + \cos 3\pi t} \right)$

When a particle executing SHM oscillates with a frequency $v$, then the kinetic energy of the particle

If a simple harmonic is represented by $\frac{{{d^2}x}}{{d{t^2}}}$$+ \alpha x = 0$,its time period is

If a watch with a wound spring is taken on to the moon, it

Out of the following functions representing motion of a particle which represents SHM (i) $y = \sin \omega t - \cos \omega t$ (ii) $y = {\sin ^3}\omega t$ (iii) $y = 5\cos \left( {\frac{{3\pi }}{4} - 3\omega t} \right)$ (iv) $y = 1 + \omega t + {\omega ^2}{t^2}$

Spring force is a