The amplitude of a damped oscillator is `A_(0)`. It becomes `(A_(0))/(3)` in 2 minutes. The amplitude of oscillation after 6 minutes will be
(A) `(A_(0))/(6)`

(B) `(A_(0))/(9)`

(C) `(A_(0))/(27)`

(D) `(A_(0))/(81)`

The maximum acceleration of a simple harmonic oscillator is a_(0) and maximum velocity is v_(0) . What is the amplitude?

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 .

A string is fixed at both end transverse oscillations with amplitude a_(0) are excited. Which of the following statements are correct ?

In damped oscillations damping froce is directly proportional to speed of ocillatior .If amplitude becomes half to its maximum value is 1 s, then after 2 s amplitude will be ( A_(0) - initial amplitude)

Let A_(0) be the area enclined by the orbit in a hydrogen atom .The graph of in (A_(0) //A_(1)) against ln(n)

If Delta=|(a_(11), a_(12), a_(13) ),(a_(21), a_(22), a_(23)),(a_(31), a_(32), a_(33))| and A_(i j) is cofactors of a_(i j) , then value of Delta is given by (A) a_(11)A_(31)+a_(12)A_(32)+a_(13)A_(33) (B) a_(11)A_(11)+a_(12)A_(21)+a_(13)A_(31) (C) a_(21)A_(11)+a_(22)A_(12)+a_(23)A_(13) (D) a_(11)A_(11)+a_(21)A_(21)+a_(31)A_(31)

Let A_(0) be the area enclined by the orbit in a hydrogen atom .The graph of in (A_(0) //A_(1)) againest in(pi)

In damped oscillations, the amplitude is reduced to one-third of its initial value a_(0) at the end of 100 oscillations. When the oscillator completes 200 oscillations ,its amplitude must be

If a_(1)gt0 for all i=1,2,…..,n . Then, the least value of (a_(1)+a_(2)+……+a_(n))((1)/(a_(1))+(1)/(a_(2))+…+(1)/(a_(n))) , is

A_(0) layer is rich in