
A transverse wave is described by the equation \[y={{y}_{0}}\sin 2\pi \left( ft-\dfrac{x}{\lambda } \right)\]. The maximum particle velocity is equal to four times the wave velocity if:
\[\begin{align}
& \text{A}\text{. }\lambda =\dfrac{\pi {{y}_{0}}}{4} \\
& \text{B}\text{. }\lambda =\dfrac{\pi {{y}_{0}}}{2} \\
& \text{C}\text{. }\lambda =\pi {{y}_{0}} \\
& \text{D}\text{. }\lambda =2\pi {{y}_{0}} \\
\end{align}\]
Answer
585.6k+ views
Hint: Find the wave velocity from the equation and remember two points on the wave always maintain a specific distance. Differentiate the wave equation to find the maximum particle velocity. Compare these two values according to the given condition to find the value of λ.
Formula used:
\[y=A\sin (\omega t-kx)\]
Complete step by step answer:
We can write a simple wave equation in the form:
\[y=A\sin (\omega t-kx)\]
Where,
\[A\] is the amplitude
\[k\] is the propagation constant
\[\omega \] is the angular velocity
Two points on the wave always maintain a specific distance. So, it retains its displacement as it moves. So, for a fixed point on the waveform, we must have a constant argument. Hence, we have
\[\omega t-kx=\]constant
We can differentiate the equation with respect to time,
\[\omega -k\dfrac{dx}{dt}=0\]
\[\Rightarrow \dfrac{dx}{dt}=\dfrac{\omega }{k}\]
\[\dfrac{\omega }{k}\] gives the wave velocity of the wave.
Hence, in this case the is given by,
\[y={{y}_{0}}\sin 2\pi \left( ft-\dfrac{x}{\lambda } \right)\].................(1)
So, we can write that,
\[v=\dfrac{f}{\left( \dfrac{1}{\lambda } \right)}=\lambda f\]
We can differentiate equation (1),
\[v={{y}_{0}}(2\pi f)\cos 2\pi \left( ft-\dfrac{x}{\lambda } \right)\]................(2)
Hence, the maximum value of the equation (2) is,
\[{{v}_{\max }}={{y}_{0}}(2\pi f)\]
It is given that,
\[{{v}_{\max }}=4v\]
So, we can write,
\[2\pi f{{y}_{0}}=4\lambda f\]
\[\Rightarrow \lambda =\dfrac{\pi {{y}_{0}}}{2}\]
Hence,
\[\lambda =\dfrac{\pi {{y}_{0}}}{2}\]
So, the correct option is (B).
Note: The quantities wave velocity and particle velocity are different for progressing waves. Wave velocity is the velocity by which the wave progresses through the medium. Wave velocity depends on the properties of the medium. It is given by,
\[\dfrac{\omega }{k}\]
Whereas, the particle velocity is a changing velocity by which the particles in the medium moves for the successful propagation of the wave. We need to differentiate the given wave equation to find the particle velocity. For example, we can differentiate the conventional form to find the velocity equation.
\[v=Aw\cos (\omega t-kx)\]
As you can see, both of these quantities are quite different from each other.
Formula used:
\[y=A\sin (\omega t-kx)\]
Complete step by step answer:
We can write a simple wave equation in the form:
\[y=A\sin (\omega t-kx)\]
Where,
\[A\] is the amplitude
\[k\] is the propagation constant
\[\omega \] is the angular velocity
Two points on the wave always maintain a specific distance. So, it retains its displacement as it moves. So, for a fixed point on the waveform, we must have a constant argument. Hence, we have
\[\omega t-kx=\]constant
We can differentiate the equation with respect to time,
\[\omega -k\dfrac{dx}{dt}=0\]
\[\Rightarrow \dfrac{dx}{dt}=\dfrac{\omega }{k}\]
\[\dfrac{\omega }{k}\] gives the wave velocity of the wave.
Hence, in this case the is given by,
\[y={{y}_{0}}\sin 2\pi \left( ft-\dfrac{x}{\lambda } \right)\].................(1)
So, we can write that,
\[v=\dfrac{f}{\left( \dfrac{1}{\lambda } \right)}=\lambda f\]
We can differentiate equation (1),
\[v={{y}_{0}}(2\pi f)\cos 2\pi \left( ft-\dfrac{x}{\lambda } \right)\]................(2)
Hence, the maximum value of the equation (2) is,
\[{{v}_{\max }}={{y}_{0}}(2\pi f)\]
It is given that,
\[{{v}_{\max }}=4v\]
So, we can write,
\[2\pi f{{y}_{0}}=4\lambda f\]
\[\Rightarrow \lambda =\dfrac{\pi {{y}_{0}}}{2}\]
Hence,
\[\lambda =\dfrac{\pi {{y}_{0}}}{2}\]
So, the correct option is (B).
Note: The quantities wave velocity and particle velocity are different for progressing waves. Wave velocity is the velocity by which the wave progresses through the medium. Wave velocity depends on the properties of the medium. It is given by,
\[\dfrac{\omega }{k}\]
Whereas, the particle velocity is a changing velocity by which the particles in the medium moves for the successful propagation of the wave. We need to differentiate the given wave equation to find the particle velocity. For example, we can differentiate the conventional form to find the velocity equation.
\[v=Aw\cos (\omega t-kx)\]
As you can see, both of these quantities are quite different from each other.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Two Planoconcave lenses 1 and 2 of glass of refractive class 12 physics CBSE

The compound 2 methyl 2 butene on reaction with NaIO4 class 12 chemistry CBSE

Bacterial cell wall is made up of A Cellulose B Hemicellulose class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

The pH of the pancreatic juice is A 64 B 86 C 120 D class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

