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 .

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 .

Equation of normal to x= 2e^(t), y= e^(-t) " at " t=0 is

If t gt0 and t^(2)-4=0 , what is the value of t ?

The half life of a radioactive element is T and its initial activity at t=0 is A_(0) and at t=t it is A, then

If the differential equation of a damped oscillator is (m d^(2)x)/(d t^(2)) + (bd x)/(d t) + k x = 0 then energy of damped oscillator vary with time .t. is E(t) =

Three light waves combine at a certain point where thcir electric field components are E_(1) = E_(0) sin omega t , E_(2)= E_(0) sin(omega t +60^(@)) , E_(3)= E_(0) sin(omega t -30^(@)) . Find their resultant component E(t) at that point.

A point performs damped oscilaltions with frequency omega and dampled coefficient beta . Find the velocity amplitude of the point as a function of time t if at the moment t=0 (a) its displacement amplitude is equal to a_(0), (b) the displacement of the point x(0)=0 and its velocity projection v_(x(0)=dot(x_(0))

The r.m.s value of Potential due to superposition of given two alternating potentials E_(1) = E_(0) sin omega t and E_(2) = E_(0) cos omega t will be

The amplitude of a particle in damped oscillation is given by A=A_0e^(-kt) where symbols have usual meanings if at time t=4s,the amplitude is half of initial amplitude then the amplitude is 1/8 of initial value t=