Answer:
Power will be 0.2023 watt
And when amplitude is halved then power will be 0.0505 watt
Step-by-step 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
![\mu =(m)/(l)=(0.0026)/(0.84)=0.0031kg/m](https://img.qammunity.org/2021/formulas/physics/college/8olwrz1c8mrsrtuif3ewgwvu4colgkbqz8.png)
Tension in the wire T = 25 N
Frequency f = 120 Hz
So angular frequency
![\omega =2\pi f=2* 3.14* 120=753.6rad/sec](https://img.qammunity.org/2021/formulas/physics/college/jw3j8jzf0o2rrm8j6zj0tll1avyeg5afbg.png)
And amplitude A = 1.6 mm = 0.0016 m
We have to find the generated power
Power is given by
![P=(1)/(2)√(\mu T)\omega ^2A^2=(1)/(2)* √(0.0031* 25)* 753.6^2* 0.0016^2=0.2023watt](https://img.qammunity.org/2021/formulas/physics/college/3hgpalf63l33v76jlrprkcf82mmwh0n4x1.png)
From the relation we can see that power
![P\ \propto\ A^2](https://img.qammunity.org/2021/formulas/physics/college/wli6u7wbcaujq0io9o7x4db3mtfh37pzco.png)
So if amplitude is halved then power will be
times
So power will be equal to
![(0.2023)/(2)=0.0505watt](https://img.qammunity.org/2021/formulas/physics/college/85akqzypz66ej403ec2885k8vcr9bpnjqk.png)