What impulse need to be given to a body of mass $m$, released from the surface of earth along a straight tunnel passsing through centre of earth, at the centre of earth, to bring it to rest(Mass of earth $M$, radius of earth R) 

A particle is suspended from a light vertical inelastic string of length 'l' from a fixed support. At its equilibrium position, it is projected horizontally with a speed $\sqrt{6gl}$. Find the ratio of tension on string, its horizontal position to that in vertically above the point of support.

The amplitude of a damped harmonic oscillator becomes halved in $\ minute$. After three minutes, the amplitude will becomes $\dfrac{1}{x}$ of initial amplitude, where $x$ is ?

A particle performing SHM is found at its equilibrium at $  t=1\ sec$ and it is found to have a speed of $0.25 \mathrm{m} / \mathrm{s}  $ at $  \mathrm{t}=2\ \mathrm{sec}  $ . If the period of oscillation is $6\ \mathrm{sec}  $. Calculate amplitude of oscillation

Assertion (A): In damped vibrations, amplitude of oscillation decreases
Reason (R): Damped vibrations indicate loss of energy due to air resistance

A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F \ sin \omega.$ If the amplitude of the particle is maximum for $\omega = \omega _1$ and the energy of the particle is maximum for $\omega = \omega _2$ then (where $\omega _0$ natural frequency of oscillation of particle)

Few particles undergo damped harmonic motion. Values for the spring constant $k$ , the damping constant $b$ , and the mass $m$ are given below. Which leads to the smallest rate of loss of mechanical energy at the initial moment?

A bar magnet oscillates with a frequency of$ 10 $ oscillations per minute. When another bar magnet is placed on its axis at a small distance, it oscillates at $14$ oscillations per minute. Now, the second bar magnet is turned so that poles are instantaneous, keeping the location same. The new frequency of oscillation will be 

The angular frequency of the damped oscillator is given by $\omega =\sqrt{\left(\frac{k}{m} -\dfrac{r^2}{4m^2}\right)}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $\dfrac{r^2}{mk}$ is $8%$, the changed in time period compared to the undamped oscillator is approximately as follows:  

The amplitude of a damped oscillator becomes $\left (\dfrac {1}{3}\right )rd$ in $2s$. If its amplitude after $6\ s$ in $\dfrac {1}{n}$ times the original amplitude, the value of $n$ is

In damped oscillations, damping force is directly proportional to speed to oscilator . If amplitude becomes half of its maximum value in 1s , then after 2 s amplitude will be (intial amplitude =$A _{0}$)

In damped oscillation mass is $1\ kg$ and spring constant $=100\ N/m$, damping coefficeint$=0.5\ kg\ s^{-1}$. If the mass displaced by $10\ cm$ from its mean position then what will be the value of its mechanical energy after $4$ seconds?

The amplitude of a damped harmonic oscillator becomes $\left (\dfrac {1}{27}\right )^{th}$ of its initial value $A _{0}$ after $6$ minute. What was the amplitude after $2\ minutes$?

The amplitude of a damped oscillator decreases to $0.9$ times its initial value in $5$ seconds. By how many times to its initial value, energy of oscillation decreases to, in $10$ seconds?

In forced oscillation displacement equation is $x(t)=A\cos(\omega _{d}t+\theta)$ then amplitude $'A'$ vary with forced angular frequency $\omega _{d}$ and natural angular frequency $'\omega'$ as (b=dumping constant)

In damped oscillation, the amplitude of oscillation is reduced to 1/3 of its initial value $A _0$ at the end of 100 oscillations. When the system completes 200 oscillations, its amplitude must be

If ${ \omega  } _{ 0 }$ is natural frequency of damped forced oscillation and p that of driving force, then for amplitude resonance

A pendulum with time of 1 s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in $s^{-1}$) is

The amplitude of a damped oscillator decreases to 0.9times its original magnitude in 5s. In another 10s it will decrease to $\alpha$ times its original magnitude, where $\alpha$ equals

A mass of 50 kg is suspended from a spring of stiffness 10 kN/m. It is set oscillating and it is observed that two successive oscillations have amplitudes of 10 mm and 1 mm. Determine the damping ratio.

A simple harmonic oscillator of angular frequency $2\ rad\ s^{-1}$ is acted upon by an external force $F = \sin t\ N$. If the oscillator is at rest in its equilibrium position at $t = 0$, its position at later times is proportional to

A body of mass $\text{600 gm}$ is attached to a spring of spring constant $\text{k = 100 N/m}$ and it is performing damped oscillations.  If damping constant is $0.2$ and driving force is $F = F _{0}$  $cos(\omega t)$  where $F _{0}=20N$  Find the amplitude of oscillation at resonance. 

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas $ \mathrm{V} _{0}  $ and its pressure is $  \mathrm{P} _{0} $ The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency.

The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be $\displaystyle\dfrac{1}{X}$ times the original, where $X$ is

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

The amplitude of a damped oscillator decreases to $0.9$ times to its original magnitude in $5s$. In another $10s$, it will decrease to $\alpha$ times to its original magnitude, where $\alpha$ equals.

A lightly damped oscillator with a frequency $\left( \omega  \right) $  is set in motion by harmonic driving force of frequency $\left( n \right) $. When $n\ll \omega $, then response of the oscillator is controlled by

On account of damping , the frequency of a vibrating body

In damped oscillations, the amplitude after $50$ oscillations is $0.8\;a _0$, where $a _0$ is the initial amplitude, then the amplitude after $150$ oscillations is

When an oscillator completes $100$ oscillations its amplitude reduces to $\displaystyle\dfrac{1}{3}$ of its initial value. What will be its amplitude when it completes $200$ oscillations?

In reality, a spring won't oscillate for ever.               will                the amplitude of oscillation until eventually the system is at rest.

Undamped oscillations are practically impossible because

Dampers are found on bridges

If we wish to represent the equation for the position of the mass in terms of a differential equation, which one of these would be the most suitable?

Two point masses $m _1$ and $m _2$ are coupled by a spring of spring. Constant $k$ and uncompressed length $L _0$. The spring is fully compressed and a thread ties the masses together with negligible separation between them. The tied assembly is moving in the $+x$ direction with uniform speed $v _0$. At a time, say $t = 0$, it is passing the origin and at that instant the thread breaks. The masses, attached to the spring, start oscillating. The displacement of mass $m _1$ given by $x _1(t) = v _0 t(1 - cos \omega t)$ where $A$ is a constant. Find (i) the displacement $x _2(t)$ is $m _2$, and (ii) the relationship between $A$ and $L _0$.

To and fro motion of a particle about its mean position is called -

The time taken by a vibrating body to complete one vibration is called its frequency. True or false.