Answer:
Power will be 0.2023 watt
And when amplitude is halved then power will be 0.0505 watt
Explanation:
We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg
Length of wire l = 84 cm = 0.84 m
So mass density [tex]\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m[/tex]
Tension in the wire T = 25 N
Frequency f = 120 Hz
So angular frequency [tex]\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec[/tex]
And amplitude A = 1.6 mm = 0.0016 m
We have to find the generated power
Power is given by [tex]P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt[/tex]
From the relation we can see that power [tex]P\ \propto\ A^2[/tex]
So if amplitude is halved then power will be [tex]\frac{1}{4}[/tex] times
So power will be equal to [tex]\frac{0.2023}{2}=0.0505watt[/tex]