Question

Two train whistles have identical frequencies of 1.80 ∗ 102 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the station platform hears beats with a frequency of 2.00 beats per second when the whistles operate together. What are the possible speeds and directions that the moving train can have?[3.85 m/s or 3.77 m/s]

Answer 1

The source speed toward the station and away from the station are 3.79 m/s and 3.87 m/s.

Step-by-step explanation:

Given that,

Frequency of the whistles f = 1.80\times10^{2}\ Hz[/tex]

Beat frequency = 2.00 b/s

We need to calculate the frequency

Using formula of beat frequency

`\Delta f=f'-f`

`f'=\delta f+f`

Put the value into the formula

`f'=2.00+180`

`f'=182.0\ Hz`

When the train moving towards station, then the frequency heard is more than the actual

We need to calculate the source speed

Using Doppler effect

`f'=f((v)/(v-v_(s)))`

`v-v_(s)=((f)/(f'))v`

Therefore, the source speed is

`v_(s)=v-v(f)/(f')`

Put the value into the formula

`v_(s)=345-345*(180)/(182)`

`v_(s)=3.79\ m/s`

When the train moving away from the station, then the frequency heard is

Again from beat frequency,

`\Delta f=f-f'`

`f'=\Delta f-f`

Put the value into the formula

`f'=180-2`

`f'=178\ Hz`

We need to calculate the source speed

Using Doppler effect

`f'=f((v)/(v+v_(s)))`

`v+v_(s)=((f)/(f'))v`

Therefore, the source speed is

`v_(s)=v(f)/(f')-v`

Put the value into the formula

`v_(s)=345*(180)/(178)-345`

`v_(s)=3.87\ m/s`

Hence, The source speed toward the station and away from the station are 3.79 m/s and 3.87 m/s.