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]

Respuesta :

Answer:

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

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

[tex]\Delta f=f'-f[/tex]

[tex]f'=\delta f+f[/tex]

Put the value into the formula

[tex]f'=2.00+180[/tex]

[tex]f'=182.0\ Hz[/tex]

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

[tex]f'=f(\dfrac{v}{v-v_{s}})[/tex]

[tex]v-v_{s}=(\dfrac{f}{f'})v[/tex]

Therefore, the source speed is

[tex]v_{s}=v-v\dfrac{f}{f'}[/tex]

Put the value into the formula

[tex]v_{s}=345-345\times\dfrac{180}{182}[/tex]

[tex]v_{s}=3.79\ m/s[/tex]

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

Again from beat frequency,

[tex]\Delta f=f-f'[/tex]

[tex]f'=\Delta f-f[/tex]

Put the value into the formula

[tex]f'=180-2[/tex]

[tex]f'=178\ Hz[/tex]

We need to calculate the source speed

Using Doppler effect

[tex]f'=f(\dfrac{v}{v+v_{s}})[/tex]

[tex]v+v_{s}=(\dfrac{f}{f'})v[/tex]

Therefore, the source speed is

[tex]v_{s}=v\dfrac{f}{f'}-v[/tex]

Put the value into the formula

[tex]v_{s}=345\times\dfrac{180}{178}-345[/tex]

[tex]v_{s}=3.87\ m/s[/tex]

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