Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and I, respectively. At time t = 0 one particle has displacement A while the other one has displacement $${{ - A} \over 2}$$ and they are moving towards each other. If they cross each other at time t, then t is :

A particle performs simple harmonic motion with amplitude $$A.$$ Its speed is trebled at the instant that it is at a distance $${{2A} \over 3}$$ from equilibrium position. The new amplitude of the motion is:

The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {{L \over g}} $$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

A pendulum made of a uniform wire of cross sectional area $$A$$ has time period $$T.$$ When an additional mass $$M$$ is added to its bob, the time period changes to $${T_{M.}}$$ If the Young's modulus of the material of the wire is $$Y$$ then $${1 \over Y}$$ is equal to :
($$g=$$ $$gravitational$$ $$acceleration$$)