Oscillations Te...

Two blocks of masses $$m_{1}=m$$ and $$m_{2}3\ m$$ are connected by a spring of force constant $$k$$ and placed on a horizontal frictionless surface as shown in the fig. The spring is stretched by an amount $$x$$ and released. The system executes simple harmonic motion. The relative velocity of the blocks when the spring is at its natural length is 

A spring executes SHM with mass of 10 kg attached to it. The force constant of spring is 10 N/m.If at any instant its velocity is 40 cm/sec. The displacement will be (here amplitude is0.5m) 

Aa particle executes simple harmonic motion with a speed of $$T\ s$$ and magnitude $$A\ m$$. The shortest time it takes to reach a point $$\dfrac{A}{\sqrt{2}}\ m$$ from its mean position in seconds is:

A pendulum clock keeps correct time at $$20^{o}C$$. The correction to be made during summer per day, when the average temperature is $$40^{o}C, (\alpha =10^{-5}/^{o}C)$$  will be:

Two pendulums have time periods $$T$$ and 5$$T / 4 .$$ They start SHM at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase:

If a simple pendulum with length $$  L  $$ and mass of the bob $$  m  $$ is vibrating with an amplitude "a', then
the maximum tension in the string is

A simple pendulum of length $$  \ell  $$ has a bob of mass $$ \mathrm{m}  $$ . The work done in projecting the bob
horizontally from the mean position to give it an angular displacement of $$  60^{\circ}  $$ is

A particle of mass m is in SHM, with velocity $$V_0$$ at its lowest position. The maximum height attained by the pendulum will be

The frequency of a simple pendulum is n oscillations per minute while that of another is (n $$+ 1 )$$ oscillations per minute. The ratio of the length of the first pendulum to the length of second is-

A particle executes S.H.M with a period of $$8s$$. Find the time in which the potential energy is equal to half of the total energy.