Questions
When a tuning fork (frequency 262 Hz) is struck, it loses half of its energy after 4s.The decay time ( ) and Q-factor for this tuning fork respectively are (Given, ln 2 = 0.693)
detailed solution
Correct option is B
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=9.5×103Talk to our academic expert!
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