For a damped oscillation m=500 gm ,k=100 N/m,b=75 gm/sec. What is the ratio of amplitue of damped oscillation to the intial amplitude at the end of 20 cycles
`["given" 1/(e^(0.675)=0.5,1/(e^0.51)=0.6)]`

For the damped oscillator of Fig. 15.20, m= 250g, k= 85N/m, and b= 70g/s What is the period of the motion?

For the damped oscillator of Fig. 15.20, m= 250g, k= 85N/m, and b= 70g/s How long does it take for the amplitude of the damped oscillations to drop to half its initial value?

What is the maximum velocity of the bob of a second's pendulum, if the amplitude of oscillation of the pendulum is 0.1 m ?

The amplitude of a lightly damped oscillator decreases by 4.0% during each cycle. What percentage of mechanical energy of the oscillator is lost in each cycle?

For a damped oscillator the mass 'm' of the block is 200 g, k = 90 N m^(-1) and the damping constant b is 40 g s^(-1) . Calculate the period of oscillation.

A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m . The magnitude of the velocity in m sec^(-1) at the centre of oscillation is

In a damped oscillation amplitude at (t = 0) is A_(0) and at (t = T) its value A_(0)//2 if E (t = 0) = E_(0) find E (t = 2T)

In dampled oscillation , the amplitude of oscillation is reduced to half of its initial value of 5 cm at the end of 25 osciallations. What will be its amplitude when the oscillator completes 50 oscillations ? Hint : A= A_(0) e^((-bt)/(2m)) , let T be the time period of oxcillation Case -I : (A_(0))/(2) = A_(0)e^(-bx(25T)/(2m)) or (1)/(2)= e^(-25(bT)/(2m)) ......(i) Case -II A=A_(0)e^(-bxx50(T)/(2m)) A _(0)(e^(-25(bT)/(2m)))^(2) Use euation (i) to find a .