Question

A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s. The observed frequency of the hum will be:

a) Greater than 152 Hz
b) Equal to 152 Hz
c) Less than 152 Hz
d) Not enough information

Answer 1

The observed frequency of the hum of a bumblebee flying past at 3 m/s will be greater than the emitted frequency of 152 Hz due to the Doppler effect, where the source is moving towards the observer.

Step-by-step explanation:

The question involves the concept of the Doppler effect, which describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source. In this case, we're dealing with a bumblebee flying past an observer at a speed of 3 m/s. According to the Doppler effect, the observed frequency of the hum will change based on the relative motion of the bumblebee to the observer.

Since sound waves are involved, the formula for the observed frequency (f') when the source is moving towards a stationary observer is:

f' = (v + vo) / (v + vs) * f

Here,

• v is the speed of sound (342 m/s),
• vo is the observer's velocity towards the source (0 m/s, because the observer is stationary),
• vs is the source's velocity towards the observer (3 m/s),
• f is the emitted frequency (152 Hz).

Inserting the values into the equation:

f' = (342 + 0) / (342 - 3) * 152

f' = 342 / 339 * 152

f' > 152 Hz

Therefore, the observed frequency is greater than the emitted frequency of 152 Hz, which corresponds to option a) Greater than 152 Hz.