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Question: In a damped oscillator with, m = 250 g, k = N/m and b = 70 g / s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 20cycles?

Short Answer

Expert verified

Answer

The ratio of oscillation amplitude at the end of 20 cycles to the initial amplitude is

x20x0=0.39

Step by step solution

01

Given

  1. The mass of the object is,m=250g=0.25kg
  2. The damping constant of the oscillator is,b=70g/s=0.070kg/s
  3. Spring constant,k=85N/m
02

Understanding the concept  

Using the equation for amplitude of forced oscillation, we can find the amplitude of oscillation after 20 cycles. Then we can find its ratio with initial amplitude.

The amplitude of oscillations after n cycles is given as-

xn=xme-bt2m

The time period of oscillation in the case of spring is given as-

T=2mk

Here, m is the mass of the pendulum, k is the force constant of the spring, b is a damping constant and t is the time taken.

03

Calculate the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 20 cycles 

Suppose the period of oscillation is T and the initial amplitude is xm

Now, we have equation for amplitude of the oscillation after n cycle is,

xn=xme-bt2m

So, for n = 20, t = 20T, we get

x20=xme-b20T2m

The time period of oscillation can be expressed as

T=2mk=20.25kg85=0.34s

Using this value of period and the given values, the amplitude of oscillation after 20 cycles is

x20=xme-0.070kg/s200.34s20.25kgx20=xm0.39

Hence, the ratio of amplitudes is

x20x0=xm0.39xm=0.39

So, the required ratio of amplitudes is 0.39.

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