JEE Main · 2005 · 08 May 2005 · Offline · Waves & Sound

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When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. If the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?

(A)200 Hz
(B)202 Hz
(C)196 Hz✓ Correct
(D)204 Hz

Answer: (C)

Solution

Before taping,

∣ f_{1} - f_{2} ∣ = 4

Hz with

f_{1} = 200

Hz, so

f_{2}

is either

196

204

Hz. Loading fork

2

with tape increases its mass and lowers its frequency, and the beats then rise to

6

Hz, meaning the two forks have moved further apart. That happens only if fork

2

already lay below fork

1

, so originally

f_{2} = 200 - 4 = 196

Hz.

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option

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