KVPY SX SB Exam 20141 Question Paper

Question : 29 of 120

Marks: +1, -0

A whistle emitting a loud sound of frequency 540 Hz is whirled in a horizontal circle of radius 2 m and at a constant angular speed o f 15 rad/s. The speed o f sound is 330 m/s.
The ratio of the highest to the lowest frequency heard by a listener standing at rest at a large distance from the center of the circle is

Validate

Solution:

v_1

and

v_2

are speed of whistle

| v_1 | = | v_2 | = ωr

= 15 × 2
= 30 m/s
Maximum frequency heard →
Here,
f → original frequency (540 Hz)
v → speed of sound

v_s

→speed of whistle

\f_\max = \f[v/{v-v_s}]

Minimum frequency heard →

\f_\min = \f[v/{v+v_s}]

\f_\max /\f_\min ={v+v_s}/{v-v_s}={330 +30}/{330 -30}

=360/300

\f_\max /\f_\min =6/5=1.2

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