Question

a girl on a swing with a period π sec and 5meters amplitude . A boy is blowing whistle with 680 hz standing at a distance of maximum frequency heard by the girl is (v=340m/sec)​

Answer 1

Answer:

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Answer 2

Given info : A girl on a swing with a period π sec and 5 metre amplitude. A boy is blowing whistle with 680 Hz at a distance.

To find : the maximum frequency heard by the girl is ....

solution : here velocity of source (the boy) = 0

velocity of observer = v₀

The maximum frequency occurs when she is moving towards the boy.

so, from Doppler's effect,

$\quad f=\left(\frac{v+v_0}{v}\right)f$

here, v is speed of sound, f is initial frequency of whistle.

given amplitude, A = 5m , Time period, T = π sec

so, v₀ = ωA = 2πA/T = 2π × 5m/πsec

= 10 m/s

now frequency heard by her, f = [(340 + 10)/340] × 680

= 350 × 2

= 700 Hz

Therefore the maximum frequency heard by the girl is 700 Hz.

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