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

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 is $${V_0}$$ and its pressure is $${P_0}.$$ 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 frquency

If a simple pendulum has significant amplitude (up to a factor of $$1/e$$ of original ) only in the period between $$t = 0s\,\,to\,\,t = \tau \,s,$$ then $$\tau \,$$ may be called the average life of the pendulum When the spherical bob of the pendulum suffers a retardation (due to viscous drag) proportional to its velocity with $$b$$ as the constant of proportionality, the average life time of the pendulum is (assuming damping is small) in seconds :

Two particles are executing simple harmonic motion of the same amplitude $$A$$ and frequency $$\omega $$ along the $$x$$-axis. Their mean position is separated by distance $${X_0}\left( {{X_0} > A} \right)$$. If the maximum separation between them is $$\left( {{X_0} + A} \right),$$ the phase difference between their motion is: