Physics Oscillations

Time period of a simple pendulum is T. The time taken to complete   $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha }{\beta }T.$  The value of a is……….

Given below are two statements:

Statement I : A second's pendulum has a time period of 1 second.

Statement II : It takes precisely one second to move between the two extreme positions.

In the light of the above statements, choose the correct answer from the options given below:

A particle executes S.H.M., the graph of velocity as a function of displacement is:

If two similar spring each of spring constant K₁ are joined in series, the new spring constant and time period would be changed by a factor

When a particle executes SHM, the nature of graphical representation of velocity as a function of

In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is:

The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as $U=\frac{\alpha }{r^{10}}-\frac{\beta }{r^5}-3$

Where, and are positive constants. The equilibrium between two atoms will be ${\left(\frac{2\alpha }{\beta }\right)}^{\frac{a}{b}},$ ${\frac{}{}}^{\frac{}{}}$ where a =……………. 

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is

Y = A sin $\left(\omega t+{?}_0\right)$ is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is y = $\frac{A}{2}$ and it moving along negative x-direction. Then the initial phase angle ${?}_0$ will be