The ends of a stretched string of length L are fixed at x-0 and x-L- In one experiment- the displacement of the wire is y1-2Asin- xLsin- t and energy E1 and in another experiment- its displacement is y2 - Asin2- x-Lsin2- t and energy E2 then

The correct option is C E_2 = 4E_1 As K_1 = πL #10;and K_2=2πL and ω_2 = 2ω_1 , #160; #10;Wave velocity v=ωK is same for # ...

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Standard XII

Physics

Transmission of Wave

The ends of a...

Question

The ends of a stretched string of length $L$ are fixed at $x = 0$ and $x = L$ . In one experiment, the displacement of the wire is $y_{1} = 2 A sin (\frac{π x}{L}) sin ω t$ and energy $E_{1}$ and in another experiment, its displacement is $y_{2} = A sin (\frac{2 π x}{L}) sin 2 ω t$ and energy $E_{2}$ then

E_{2} = E_{1}

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E_{2} = 2 E_{1}

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E_{2} = 4 E_{1}

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16 E_{1}

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Solution

The correct option is C $E_{2} = 4 E_{1}$
As $K_{1} = \frac{π}{L}$ and $K_{2} = \frac{2 π}{L}$ and $ω_{2} = 2 ω_{1}$ ,
Wave velocity $v = \frac{ω}{K}$ is same for both the cases as well as amplitudes in both the cases .
Hence , Energy $\propto f^{2} \to$ Energy $\propto ω^{2}$
$E_{2} = 4 E_{1}$

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The ends of a stretched string of length L are fixed at x=0 and x=L. In one experiment, the displacement of the wire is y1=2Asin(πxL)sinωt and energy E1 and in another experiment, its displacement is y2=Asin(2πxL)sin2ωt and energy E2 then

The correct option is C E2=4E1As K1=πL and K2=2πL and ω2=2ω1 , Wave velocity v=ωK is same for both the cases as well as amplitudes in both the cases . Hence , Energy ∝f2→ Energy ∝ω2 E2=4E1