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NVAEasyJEE 2023Doppler Effect

JEE Physics 2023 Question with Solution

A train blowing a whistle of frequency 320 Hz320 \, \text{Hz} approaches an observer standing on the platform at a speed of 66 m/s66 \, \text{m/s}. The frequency observed by the observer will be (given speed of sound = 330 m/s330 \, \text{m/s}):

Answer

Correct answer:400

Step-by-step solution

Standard Doppler Effect Method

Given: Source frequency f=320 Hzf = 320 \, \text{Hz}, speed of source vs=66 m/sv_s = 66 \, \text{m/s}, speed of sound v=330 m/sv = 330 \, \text{m/s}.

Find: The observed frequency when the source approaches a stationary observer.

A train on a track moving toward a stationary observer on the platform, with speed labeled 66 m/s and a whistle shown on the train.

For a source approaching a stationary observer, the Doppler effect formula is

fapp=f(vv−vs)f_{\text{app}} = f \left(\frac{v}{v-v_s}\right)

Substituting the given values,

fapp=320(330330−66)f_{\text{app}} = 320\left(\frac{330}{330-66}\right) fapp=400 Hzf_{\text{app}} = 400 \, \text{Hz}

Therefore, the observed frequency is 400 Hz400 \, \text{Hz}.

Equivalent Fraction Form

Given: The whistle frequency is 320 Hz320 \, \text{Hz} and the train approaches the observer with speed 66 m/s66 \, \text{m/s}.

Find: The frequency heard by the observer.

Using the Doppler effect relation for an approaching source,

f′=f1−vsvsoundf' = \frac{f}{1 - \frac{v_s}{v_{\text{sound}}}}

Substituting the values,

f′=320 Hz1−66 m/s330 m/sf' = \frac{320 \, \text{Hz}}{1 - \frac{66 \, \text{m/s}}{330 \, \text{m/s}}} f′=400 Hzf' = 400 \, \text{Hz}

Hence, the final answer is 400400.

Common mistakes

  • Using the receding-source Doppler formula is incorrect because the train is moving toward the observer. For an approaching source, the denominator becomes v−vsv - v_s, which increases the observed frequency.

  • Treating the observer as moving is wrong here because the observer is standing on the platform. Use the stationary-observer form of the Doppler formula, not the moving-observer expression.

  • Subtracting the source speed directly from the frequency is incorrect because Doppler effect depends on the ratio of wave speeds, not on simple arithmetic difference of frequencies. Substitute into the formula before calculating.

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