Question
Question: A transverse wave is described by the equation \(y={y}_{0}\sin {2 \pi} (ft- \dfrac{\pi}{\lambda})\)....
A transverse wave is described by the equation y=y0sin2π(ft−λπ). The maximum particle velocity is equal to the four times the wave velocity if:
A. λ=4πy0
B. λ=2πy0
C. λ=πy0
D. λ=2πy0
Solution
To solve this problem, first find the maximum particle velocity. To find the maximum particle velocity derive the given equation of transverse wave. Then find the wave velocity. Use the condition given as the maximum particle velocity is equal to the four times wave velocity. Substitute the equations obtained above. Rearrange the equation and find the wavelength. For this obtained wavelength, the maximum particle velocity will be four times the wave velocity.
Complete answer:
Let the maximum velocity of the particle be vmax
The wave velocity be vwave
A transverse wave is described by the equation,
y=y0sin2π(ft−λπ)...(1)
Particle velocity is given by,
v=dtdy
So, differentiating the equation. (1) we get,
v=y0fcos2π(ft−λπ)
For maximum particle velocity,
cos2π(ft−λπ)=1
Thus, the maximum particle velocity will be given by,
vmax=y02πf
Velocity of the wave is given by,
vwave=λf
The condition is given as,
vmax=4vwave
Substituting the values in above condition we get,
y02πf=4λf
⇒y0π=2λ
⇒λ=2y0π
Hence, the maximum particle velocity is equal to the four times the wave velocity of 2y0π.
So, the correct answer is “Option B”.
Note:
Students must remember the difference between particle velocity and wave velocity. They should not get confused between them. Both these quantities are different for progressive waves. Wave velocity is the velocity of a particle by which id=t is progressing through the medium. Whereas, the particle velocity is the changing velocity of the particle when it travels from successful propagation of the wave, Velocity of the particle at the instant of passing through the mean position is maximum. While it is minimum at the extreme positions.