Question

Discuss the following case:

Source in motion and Observer at rest

1. Source moves towards observer
2. Source moves away from the observer

Answer 1

(a) Source moves towards die observer-

Source S moves towards an observer O (right) with velocity

Suppose a source S moves to the right (as shown in Figure) with a velocity v_s and let the frequency of the sound waves produced by the source be f_s. It is assumed that the velocity of sound in a medium is v.

Suppose a source S moves to the right (as shown in Figure) with a velocity v_s and let the frequency of the sound waves produced by the source be f_s. It is assumed that the velocity of sound in a medium is v.

The compression (sound wavefront) produced by the source S at three successive instants of time are shown in the Figure. When S is at position x₁ the compression is at C₁. When S is at position x₂, the compression is at C₂ and similarly for x₃ and C₃.

It is meant that the wavelength decreases when the source S moves towards the observer O. But frequency is inversely related to wavelength and therefore, frequency increases.

Calculation:

Let λ be the wavelength of the source S as measured by the observer when S is at position x₁ and λ’ be the wavelength of the source observed by the observer when S moves to position x₂.

Then the change in wavelength is ∆λ = λ – λ’ = v_st, where t is the time taken by the source to travel between x₁ and x₂ Therefore,

λ’ = λ – v_st … (1)

But t = \[\frac{λ}{v}\] … (2)

On substituting equations (2) in equation (1), we get.

`lambda' = lambda(1 - "v"_"s"/"v")`

Since frequency is inversely proportional to wavelength, we have

`"f"' = "v"_"s"/(lambda')` and f = `"v"_"s"/lambda`

Hence `"f"' = "f"/((1 - "v"_"s"/"v"))`   ...(3)

Since, `"v"_"s"/"v"` << 1, by using the binomial expansion and retaining only first order in `"v"_"s"/"v"`, we get

`"f"' = "f"(1 + "v"_"s"/"v")`   ...(4)

(b) Source moves away from the observer-

Since the velocity of the source is opposite in direction when compared to case (a), hence by changing the sign of the velocity of the source in the above case i.e., by substituting (v_s → – v ) in equation (1), we get

`"f"' = "f"/((1 + "v"_"s"/"v"))`    ....(5)

Using binomial expansion again, we get

`"f"' = "f"(1 - "v"_"s"/"v")`   .....(6)