If the amplitude of a transverse wave is doubled then average rate of transmission of potential energy becomes
A
one fourth of initial value
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B
twice of its initial value
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C
four times of its initial value
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D
none of these
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Solution
The correct option is C four times of its initial value We know that
Average rate of transmission of potential energy is, <dUdt>=14μvω2A2 ⇒<dU′dt>=14μvω2A′2
Given: A′=2A ⇒<dU′dt>=14μvω2(2A)2 ⇒<dU′dt>=4(14μvω2A2) ∴<dU′dt>=4<dUdt>
Hence average rate of transmission of potential energy becomes 4times of its initial value.