Question

A submarine heads toward a port, which broadcasts a signal at 175 MHz. If the submarine heads toward the port at 46 m/s, what change in frequency does the submarine observe in the port's signal

Answer 1

Answer:

Δf = 0.00003 MHz = 30 Hz

Explanation:

The apparent frequency due to the motion of receiver towards the source is given by Doppler's effect, while the source is stationary. Therefore, the formula is given as follows:

$f = (\frac{c+v_{o} }{c})f_{0}$

where,

f = apparent frequency

c = speed of light = 3 x 10⁸ m/s

v₀ = velocity of observer = 46 m/s

f₀ = Original Frequency = 175 MHz = 175 x 10⁶ Hz

Therefore,

$f = (\frac{300000000 m/s + 46 m/s}{300000000 m/s})(175 MHz)$

f = 175.00003 MHz

Therefore, the change in frequency is:

Δf = f - f₀

Δf = 175.00003 MHz - 175 MHz

Δf = 0.00003 MHz = 30 Hz