A particle executing S.H.M. Its displacement x varies with time t as $x=8\sin 4t+6\cos 4t.$  The maximum  velocity of the particle will be:

The amplitude of oscillation for a particle executing S.H.M with angular frequency $\mathrm{\pi }$rad/s and maximum acceleration $5{\mathrm{\pi }}^2 \mathrm{m}/{\mathrm{s}}^2$ is

Which of the following statements is true for an ideal simple pendulum?

A mass of 30 gm is attached with two springs having spring constants 100 N/m and 200 N/m and other ends of springs are attached to rigid walls as shown in the given figure. The angular frequency of oscillation is: (Ground is smooth)

If a bob of the simple pendulum having negative charge q is made to oscillate in a uniform electric field acting vertically upward direction, then its time period:

A simple pendulum of length l and bob of mass m is executing S.H.M with amplitude A. The maximum tension in the string will be

Find the percentage change in time period if the length of a simple pendulum is increased by 3% 

A disc is oscillating about a horizontal axis passing through its rim and perpendicular to the plane of the disc. If the radius of the disc is R, then the frequency of oscillation is

The amplitude of a damped oscillator becomes one-third in 10 minutes and $\frac{1}{\mathrm{n}}$ times of the original value in 30 minutes. The value of n is:

A particle is executing SHM with time period T. The time taken by it to travel from mean position to $\frac{1}{\sqrt{2}}$ times its amplitude is equal to

A block of mass 2 kg is hanging with a massless spring and the spring is stretched by 40 cm. If the block is pulled down and released, then the period of oscillation is: (here, g = 10 $\mathrm{m}/{\mathrm{s}}^2$)

A spring is attached vertically to the ceiling of a lift and the lower end of spring is connected with a block of mass 2 kg. If the lift starts accelerating upwards with an acceleration $2 \mathrm{m}/{\mathrm{s}}^2$, then find the amplitude of S.H.M, while spring constant is 100 N/m

A block of mass m attached to a spring of constant k oscillates on a smooth surface. The other end of the spring is fixed to a wall. When spring is at its natural length, the speed of the block is v. Displacement of the block from its mean position before coming to instantaneous rest is:

The motion of the particle is started at t = 0 and the equation of motion is given by $x = 8 \sin \left(100t + \frac{\mathrm{\pi }}{6}\right)$, where x is in cm and t is in seconds. When will the particle come to rest for the first time?

What will be the frequency of oscillation of a simple pendulum, if the length of the pendulum is equal to the radius of earth?

The curve between the potential energy (U) and displacement (x) is shown. Which of the oscillation is about the mean position x =0?

The time period of a body under S.H.M. is ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$ when restoring forces ${\mathrm{F}}_1$ and ${\mathrm{F}}_2$ respectively act on it. What will be the time period of S.H.M. when both the forces act simultaneously on it?

A particle is subjected to two mutually perpendicular SHM such that x = 2sin$\mathrm{\omega }$t and

$\mathrm{y} = 2 \sin \left[\mathrm{\omega t} + \frac{\mathrm{\pi }}{2}\right]$. The path of the particle will be

Suppose that a pendulum clock is carried to a depth of 32 km inside the earth (Radius R = 6400 km). To have the correct time from the clock, by what percentage the effective length of the pendulum should be changed? 

When a periodic force $\overrightarrow{{\mathrm{F}}_1}$ acts on a particle, the particle oscillates according to the equation x=Asin$\mathrm{\omega }$t. Under the effect of another periodic force $\overrightarrow{{\mathrm{F}}_2}$, the particle oscillates according to the equation y=Bsin($\mathrm{\omega }$t+$\mathrm{\pi }$/2). The amplitude of oscillation when the force ($\overrightarrow{{\mathrm{F}}_1}$ + $\overrightarrow{{\mathrm{F}}_2}$) acts are:

A particle is executing SHM with amplitude A and angular frequency $\mathrm{\omega }$. The time taken by the particle to move from x = 0 to x = A/2 is

A spring-block system oscillates with time period T on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes T'. Then one can conclude that:

A particle executes simple harmonic oscillations under the effect of small damping. If the amplitude of oscillation becomes half of the initial value of 16 mm in five minutes, then what will be the amplitude after fifteen minutes?

Which of the following is incorrect about simple harmonic oscillations of a particle? 

For a particle executing SHM, the graph between its momentum and displacement will be

A particle executes SHM of amplitude 10 m and period 4 s along a straight line. The velocity of the particle at a distance of 6 m from the mean position is: 

A uniform rod of mass m and length L pivoted from its one end is executing SHM with time period T. If rod suddenly breaks from the middle while passing through its mean position, then the time period of oscillation of remaining half part will be

A pendulum oscillates about its mean position C. The position where the speed of the bob becomes maximum is

A particle is executing linear S.H.M. between x = A. The time taken to go from 0 to A/2 is ${\mathrm{T}}_1$ and to go from A/2 to A is ${\mathrm{T}}_2$, then

The position x (in centimeter) of a simple harmonic oscillator varies with time t (in second) as $x = 2\cos \left(0.5\mathrm{\pi t} + \frac{\mathrm{\pi }}{3}\right)$. The magnitude of the maximum acceleration of the particle in $\mathrm{cm}/{\mathrm{s}}^2$ is: