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A sinusoidal wave of frequency 500 Hzhas a speed of 350 m/s . (a)How far apart are two points that differ in phase byπ/3rad ? (b)What is the phase difference between two displacements at a certain point at times 1.00 msapart?

Short Answer

Expert verified

a) The length between two points that differ in phase by π3is 117 mm .

b) The phase difference between two displacements at a certain point at times 1.00 ms apart is πrad.

Step by step solution

01

The given data

  • The frequency of the wave,f=500Hz
  • The speed of the wave,v=350m/s
  • The time period of the wave, T=1.00ms
02

Understanding the concept of wave equation

The number of oscillations completed by the wave, per unit time interval is known as frequency. The distance between two consecutive crest or trough gives the wavelength of the wave. The product of wavelength of the wave and the frequency gives us the speed of the propagation of wave.

Formula:

The speed of the wave, v=fλ (i)

The frequency of the wave, f=1T (ii)

03

a) Calculation of the length between two points

Using equation (i), we can find the length of one cycle of the wave as given:

λ=vf=350m/s500s-1=0.7m=700mm

A cycle is equivalent to 2πradso that π3 corresponds to one-sixth of a cycle.

The corresponding length, therefore is,

λ6=700mm6=116.67mm≈117mm

Hence, the value of distance between the two points is 117 mm

04

b) Calculation of phase difference between two displacements

From equation (ii), we can find the time for one cycle of oscillation as given:

T=1f=1500s-1=2.00×10-3s=2.00ms

The interval 1.00 ms is half of T , thus corresponding to half of one cycle. Thus, the phase difference is given by

12×2π=πrad

Hence, the value o the phase difference isπrad .

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