A tuning fork has a frequency of 300 Hz and hence a period of 3.33 × 10−³ s.If the tuning fork is in an inertial frame of reference moving by the observer at speed of 0.620 c,what is the frequency of the fork as measured by the observer? (Assume that measurements are strictly by optical means and that the speed of sound waves in air is not pertinent here). A)235 Hz B)382 Hz C)185 Hz D)185 Hz. E)487 Hz

Question

Quizplus · Accepted Answer

This is a problem of relativistic Doppler effect. We need to use the formula:
f' = fγ(1 ± v/c)
where f is the frequency of the source (tuning fork), f' is the frequency observed by the observer, v is the relative velocity (0.620 c), c is the speed of light, and γ is the Lorentz factor given by:
γ = 1/√(1 - v^2/c^2)

Substituting the given values, we get:
γ = 1/√(1 - (0.620 c)^2/c^2) = 1.5
f' = 300 × 1.5(1 + 0.620) = 300 × 2.79 = 837 Hz (when moving towards the observer)
or
f' = 300 × 1.5(1 - 0.620) = 300 × 0.705 = 211.5 Hz (when moving away from the observer)

Since the tuning fork is moving towards the observer, the frequency observed will be higher than the actual frequency, and the correct answer is the option A with a frequency of 235 Hz.

A Tuning Fork Has a Frequency of 300 Hz and Hence