Find the power output of the osciallator - Please read the whole question

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To find the power output of the oscillator connected to a steel wire under tension, the relevant equation is P = 1/2 (m/L) (2πf)² A² v. To calculate m/L, the mass per unit length, the density of steel and the diameter of the wire must be used. Given the tension of 6.3 N and frequency of 63.0 Hz, the amplitude of the wave is 0.50 cm. If the frequency is doubled while keeping power constant, the new amplitude can be determined using the same power equation. Understanding these relationships is crucial for solving the problem accurately.
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A small steel wire of diameter 1.4 mm is connected to an oscillator and is under a tension of 6.3 N . The frequency of the oscillator is 63.0 Hz and it is observed that the amplitude of the wave on the steel wire is 0.50 cm.

a)What is the power output of the oscillator, assuming that the wave is not reflected back?

b)If the power output stays constant but the frequency is doubled, what is the amplitude of the wave?

Which equation should I use?

Is the equation: P = 1/2 (m/L) (2pi f)^2 A^2 v correct?
If yes, how do I find m/L, f, A, and v?

Please help me to get started on this problem
 
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