
A taut string for which $\mu =5.00\times {{10}^{-2}}kg/m$ is under a tension of 80.0 N. How much power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0 Hz and an amplitude of 6.00 cm?
Answer
232.8k+ views
Hint:To find the power needed to generate sinusoidal wave in a taut string, we first find total energy for sinusoidal wave associated with the string using the formula $E=\dfrac{1}{2}\mu x{{\omega }^{2}}{{A}^{2}}$ and then take a differentiation of it with respect to time to find power.
Formula used:
$E=\dfrac{1}{2}\mu x{{\omega }^{2}}{{A}^{2}}$
Complete step-by-step answer:
For a sinusoidal wave –
Total energy is given by $E=\dfrac{1}{2}\mu x{{\omega }^{2}}{{A}^{2}}$
Where,
$\mu $ is the linear mass density of string
x is the length element of string
$\omega $ is the angular frequency of wave, and A is the amplitude
So, power will be given by
$P=\dfrac{dE}{dt}$
After putting value of Energy (E) into power formula we get
$P=\dfrac{1}{2}\mu {{\omega }^{2}}{{A}^{2}}v$
Where, v will be the speed of the sinusoidal wave, i.e., $\dfrac{dx}{dt}$ along the string.
Velocity can be found by $v=\sqrt{\dfrac{T}{\mu }}$, where T is the tension applied on string.
We want to find the power needed to generate sinusoidal waves of frequency 60 Hz and 6 cm amplitude.
At first, we find the velocity of wave along string, which is $v=\sqrt{\dfrac{T}{\mu }}$
Substituting the values, we get
$v=\sqrt{\dfrac{80}{5\times {{10}^{-2}}}}=40m/\sec $
Now we need to convert frequency into angular frequency
$\begin{align}
& \omega =2\pi \theta \\
& \Rightarrow \omega =2\pi \times 60=377{{\sec }^{-1}} \\
\end{align}$
Now it is given the amplitude is 6cm, converting it to metre, we get
A = 0.06m
Substituting these values into the formula of power, we get
$\begin{align}
& P=\dfrac{1}{2}\mu {{\omega }^{2}}{{A}^{2}}v \\
& \Rightarrow P=\dfrac{1}{2}\times \left( 5\times {{10}^{-2}} \right)\times {{\left( 377 \right)}^{2}}\times {{\left( 0.06 \right)}^{2}}\times 40 \\
& \Rightarrow P=512W \\
\end{align}$
Hence, 512W power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0 Hz and an amplitude of 6.00 cm
Note: To generate sinusoidal waves in a string, we generate an impulse at one end of it, which travels along the string.
These kinds of waves generally need a medium to travel, in this question wave is travelling through a string.
Power is generally defined as the rate at which work is done. So mathematically it is a time differential of Energy or work done.
In the above question, our aim is to find the power required to generate a sinusoidal wave along a string which is under tension. We first find the total energy contained in this wave and take a time differentiation of it to find the power. After putting all the values in the power formula, we find the numerical value of it.
Formula used:
$E=\dfrac{1}{2}\mu x{{\omega }^{2}}{{A}^{2}}$
Complete step-by-step answer:
For a sinusoidal wave –
Total energy is given by $E=\dfrac{1}{2}\mu x{{\omega }^{2}}{{A}^{2}}$
Where,
$\mu $ is the linear mass density of string
x is the length element of string
$\omega $ is the angular frequency of wave, and A is the amplitude
So, power will be given by
$P=\dfrac{dE}{dt}$
After putting value of Energy (E) into power formula we get
$P=\dfrac{1}{2}\mu {{\omega }^{2}}{{A}^{2}}v$
Where, v will be the speed of the sinusoidal wave, i.e., $\dfrac{dx}{dt}$ along the string.
Velocity can be found by $v=\sqrt{\dfrac{T}{\mu }}$, where T is the tension applied on string.
We want to find the power needed to generate sinusoidal waves of frequency 60 Hz and 6 cm amplitude.
At first, we find the velocity of wave along string, which is $v=\sqrt{\dfrac{T}{\mu }}$
Substituting the values, we get
$v=\sqrt{\dfrac{80}{5\times {{10}^{-2}}}}=40m/\sec $
Now we need to convert frequency into angular frequency
$\begin{align}
& \omega =2\pi \theta \\
& \Rightarrow \omega =2\pi \times 60=377{{\sec }^{-1}} \\
\end{align}$
Now it is given the amplitude is 6cm, converting it to metre, we get
A = 0.06m
Substituting these values into the formula of power, we get
$\begin{align}
& P=\dfrac{1}{2}\mu {{\omega }^{2}}{{A}^{2}}v \\
& \Rightarrow P=\dfrac{1}{2}\times \left( 5\times {{10}^{-2}} \right)\times {{\left( 377 \right)}^{2}}\times {{\left( 0.06 \right)}^{2}}\times 40 \\
& \Rightarrow P=512W \\
\end{align}$
Hence, 512W power must be supplied to the string to generate sinusoidal waves at a frequency of 60.0 Hz and an amplitude of 6.00 cm
Note: To generate sinusoidal waves in a string, we generate an impulse at one end of it, which travels along the string.
These kinds of waves generally need a medium to travel, in this question wave is travelling through a string.
Power is generally defined as the rate at which work is done. So mathematically it is a time differential of Energy or work done.
In the above question, our aim is to find the power required to generate a sinusoidal wave along a string which is under tension. We first find the total energy contained in this wave and take a time differentiation of it to find the power. After putting all the values in the power formula, we find the numerical value of it.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

