For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of 4 cm from the mean position is $1 / 3$ rd of the maximum kinetic energy. The amplitude of the motion is

A block is in simple harmonic motion (SHM) on the end of the spring with position given by $x=5 \cos \left(\omega t+\frac{\pi}{4}\right) \mathrm{cm}$. If the total mechanical energy used is 100 J to achieve maximum displacement, then the potential energy at time, $t=0$ is

A particle performs simple harmonic motion with a time period of 16 s . At a time $t=2 \mathrm{~s}$, the particle passes through the origin and at $t=4 \mathrm{~s}$ its velocity is $4 \mathrm{~m} / \mathrm{s}$. The amplitude of the motion is

The amplitude of a damped oscillator varies with time as $A(t)=A_0 \exp (-b t / 2 \mathrm{~m})$, where $b=70 \mathrm{~g} / \mathrm{s}$ and $m=200$ g. How long does it take for the mechanical energy to drop to one-fourth of its initial value?

[Take, $\ln 2=0.7$ ]