A block of mass 100 ~g is connected to an elastic spring of spring constant 450 Nm ⁻¹ oscillates vertically. The block-spring system is in viscous surrounding medium with a damping constant 69.3 ~g ~s ⁻¹ . The time in which the amplitude of oscillations drop to half of its initial value. (take, 2=0.693)

Step-by-step solution

1 ) Amplitude of a damped oscillator varies with time as A=A_ 0 . e^ - t Here, =b / 2 m, b= damping constant and m= mass of oscillator. Here, given A=A₀ / 2 So, from eq. (i) we have aligned & A₀ 2 =A₀ e^ - t & 1 2 =e^ - t & ( 1 2 )= (e^ - t ) - 2=- t & or & 2= t & =b / 2 m where， & b=69.3 gs ⁻¹ and m=100 ~g & aligned Now, =b / 2 m where, and 2=0.693 (given) Substituting given values in eq. (ii), we have aligned & 0.693= 69.3 2 100 t t & =2 ~s aligned