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Q.

When a tuning fork (frequency 262 Hz) is struck, it loses half of its energy after 4s.The decay time ( τ) and Q-factor for this tuning fork respectively are (Given, ln 2 = 0.693)

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a

4.77s,10.5×103

b

5.77s,9.5×103

c

6.77s,11.5×103

d

2.77s,8.5×103

answer is B.

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Detailed Solution

We know E=E0e−t/τE=E02∴E02=E0e−t/τ⇒e4/τ=2Taking natural logarithms on both the sides 4τ=ln2⇒τ=4ln2=5.77sNow, Q=ω0τ=(2πf)τ=2×3.14×262×5.77=9.5×103

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When a tuning fork (frequency 262 Hz) is struck, it loses half of its energy after 4s.The decay time ( τ) and Q-factor for this tuning fork respectively are (Given, ln 2 = 0.693)