This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 9 Jan 2019 Shift 2)

Option 1 : 666 Hz

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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90 Questions
360 Marks
180 Mins

**Concept:**

Most vibrating objects have more than one resonant frequency and those used in musical instruments typically vibrate at harmonics of the fundamental. A harmonic is defined as an integer multiple of the fundamental frequency.

The Doppler Effect (or the Doppler shift) is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.

From question, an open flute produces second harmonic sound waves.

The length and wavelength relation are given by the formula:

L = λ_{2}

We know that, λ = v/f

\(\Rightarrow {\rm{L}} = \frac{{\rm{v}}}{{\rm{f}}}\)

\(\Rightarrow {\rm{f}} = \frac{v}{L}\)

Where,

L = Length of the flute = 50 × 10^{-2} m

v = Velocity of wave = 330 m/s

f = Frequency of second harmonic wave

λ_{2} = Wavelength of second harmonic wave

**Calculation:**

Now,

\(\Rightarrow {\rm{f}} = \frac{{330}}{{50{\rm{\;}} \times {\rm{\;}}{{10}^{ - 2}}}}\)

∴ f = 660 Hz

Now, a person runs towards the musician from another end of a hall,

v_{o} = 10 km/hr

Here, Doppler Effect is taking place.

The relationship between emitted frequency and received frequency is given by the formula:

\({f_o} = f\left[ {\frac{{v + {v_o}}}{v}} \right]\)

Where,

f_{o} = Frequency of heard by the observer

Now,

\(\Rightarrow {f_o} = 660\left[ {\frac{{330 + \left( {\frac{{10 \times {{10}^3}}}{{60 \times 60}}} \right)}}{{330}}} \right]\)

⇒ f