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A damped harmonic oscillator consists of a block (m=2.00kg), a spring (k=10.0N/m), and a damping force (F=-bv). Initially, it oscillates with amplitude of25.0cm; because of the damping, the amplitude falls to three-fourths of this initial value at the completion of four oscillations. (a) What is the value of b? (b) How much energy has been 鈥渓ost鈥 during these four oscillations?

Short Answer

Expert verified
  1. The value of b is 0.102 kg/s.
  2. The energy lost during 4 oscillationsis 0.137J .

Step by step solution

01

The given data

  • The spring constant is, k=10N/m.
  • Mass of the block isM=2.00kg.
  • The amplitude of oscillation is,xm=25cmor0.25m.
  • Because of damping, x=34xm.
02

Understanding the concept of damped oscillations

We can find the damping constant using the formula for the displacement of the damped oscillator and period of SHM. Using this constant and damping amplitude, we can find the energy at t=4T. From this, we can find the energy lost during 4 oscillations.

Formulae:

The displacement equation of the damped oscillations,x=xme-bt2m(i)

The period of oscillation for SHM,T=2mk(ii)

The potential energy of the system, (PE)=12kx2(iii)

03

a) Calculation of the damped constant

Using the displacement equation (i), we get

e-bt2m=34(x=34xm)

Since,t=4T,

e-b(4T)2m=34(iv)

The period of oscillation for SHM using equation (ii)is given as:

T=2210=2.81s

Then, substituting the value of time period in equation (iv), we get

e-b(42.81)2m=34b(42.81)2m=In43b=In43(2m)11.24=0.288(2)(2)11.24=0.102kg/s

Therefore, the value of b is 0.102 kg/s .

04

b) Calculation of lost energy during 4 oscillations

The initial energy of the system is,

E=P.E=12kx2m=12(10.0N/m)(0.250m)2 =0.313J

At role="math" localid="1657268926058" t=4T,the energy of the damped system is given as;

E'=12k34xm2(x=34xm)=916(0.313)J=0.176J

The energy lost during 4 oscillations is,

E-E'=0.313J-0.176J=0.137J

Therefore, the energy lost during 4 oscillations is 0.137 J .

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