The ends of a stretched wire of length are fixed at and . In one experiment, the displacement of the wire is and energy is and in another experiment its displacement is and energy is . Then [2001]

Question

The ends of a stretched wire of length $L$ are fixed at $x = 0$ and $x = L$ . In one experiment, the displacement of the wire is $y_{1} = A \sin (\frac{π x}{L}) \sin ω t$ and energy is $E_{1}$ and in another experiment its displacement is $y_{2} = A \sin (\frac{2 π x}{L}) \sin 2 ω t$ and energy is $E_{2}$ . Then [2001]

Smart Achievers · Accepted Answer

(3)

$Energy, E \propto A^{2} ν^{2} where A = amplitude and ν = frequency.$

$Also ω = 2 π ν \Rightarrow ω \propto ν$

$Experiment 1: Amplitude = A and ν_{1} = ν$

$Experiment 2: Amplitude = A and ν_{2} = 2 ν$

$\frac{E_{2}}{E_{1}} = \frac{A^{2} ν_{2}^{2}}{A^{2} ν_{1}^{2}} = 4$

$∴$ $E_{2} = 4 E_{1}$

Solution

1 $E_{2} = E_{1}$

2 $E_{2} = 2 E_{1}$

3 $E_{2} = 4 E_{1}$

4 $E_{2} = 16 E_{1}$

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Solution

Q. The ends of a stretched wire of length L are fixed at x=0 and x=L. In one experiment, the displacement of the wire is y1=Asin(πxL)sinωt and energy is E1 and in another experiment its displacement is y2=Asin(2πxL)sin2ωt and energy is E2. Then [2001]

1 E2=E1

2 E2=2E1

3 E2=4E1

4 E2=16E1

Ans. (3) Energy, E∝A2ν2 where A=amplitude and ν=frequency. Also ω=2πν ⇒ ω∝ν Experiment 1: Amplitude =A and ν1=ν Experiment 2: Amplitude =A and ν2=2ν E2E1=A2ν22A2ν12=4 ∴ E2=4E1

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1 $E_{2} = E_{1}$

2 $E_{2} = 2 E_{1}$

3 $E_{2} = 4 E_{1}$

4 $E_{2} = 16 E_{1}$