Question

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is given by -b(dx/dt), where b = 210 g/s. The block is pulled down 13.0 cm and released. (a) Calculate the time required for the amplitude of the resulting oscillations to fall to 1/8 of its initial value. (b) How many oscillations are made by the block in this time?

Answer 1

a) t=24s

b) number of oscillations= 11

Step-by-step explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀
`e^{(-bt)/(2m)}cos(w't+\phi)=A(t)cos(w't+\phi)`

where A(t)=A₀
`e^{(-bt)/(2m)}`

A₀ is the amplitude at t=0 and

`w'` is the angular frequency of damped SHM, which is given by,

`w'=\sqrt{(k)/(m)-(b^(2))/(4m^(2)) }`

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀
`e^{(-bt)/(2m)}` =A₀/8

⇒
`e^{(bt)/(2m)}=8`

applying logarithm on both sides

⇒
`(bt)/(2m)=ln(8)`

⇒
`t=(2m*ln(8))/(b)`

substituting the values

`t=(2*1.2*ln(8))/(0.21)=24s(approx)`

b)
`w'=\sqrt{(k)/(m)-(b^(2))/(4m^(2)) }`

`w'=\sqrt{(9.8)/(1.2)-(0.21^(2))/(4*1.2^(2))}=2.86s^(-1)`

`T'=(2\pi)/(w')`, where
`T'` is time period of damped SHM

⇒
`T'=(2\pi)/(2.86)=2.2s`

let
`n` be number of oscillations made

then,
`nT'=t`

⇒
`n=(24)/(2.2)=11(approx)`