If the maximum speed of a particle in a medium carrying a traveling wave is V_0, then find the speed of the particle when displacement is half of maximum value.

To solve the problem, let's analyze the relationship between the speed and displacement of a particle in a traveling wave.
Given:
• Maximum speed of the particle = V0V_0,
• Displacement of the particle = half of the maximum displacement.
We need to find the particle's speed vv at this displacement.
Key Equations:
The equation for the displacement yy of a particle in a traveling wave is:
y=Asin⁡(ωt)y = A \sin(\omega t)
where:
• AA is the amplitude (maximum displacement),
• ω\omega is the angular frequency,
• tt is time.
The velocity of the particle is the time derivative of displacement:
v=dydt=Aωcos⁡(ωt)v = \frac{dy}{dt} = A \omega \cos(\omega t)
At maximum velocity V0V_0, the displacement is zero (y=0y = 0):
V0=AωV_0 = A \omega
At any displacement yy, the relationship between velocity and displacement is:
v=±(Aω)2−(ωysin⁡(ωt))2v = \pm \sqrt{(A \omega)^2 - \left(\frac{\omega y}{\sin(\omega t)}\right)^2}
Velocity When Displacement Is Half the Maximum:
Let the displacement be y=A2y = \frac{A}{2}. Substituting this into the velocity equation:
v=±(Aω)2−(ωAsin⁡(y)))2v = \pm \sqrt{(A \omega)^2 - \left(\frac{\omega A}{\sin(y))}\right)^2}
Let's refine the explanation to determine the speed when the displacement is half the maximum value.
Analysis:
The total energy of the particle in the wave is conserved. This energy consists of two parts:
1. Kinetic energy (due to the velocity),
2. Potential energy (due to displacement).
At maximum displacement (y=Ay = A), all energy is potential, and the velocity is zero. At equilibrium (y=0y = 0), all energy is kinetic, and the velocity is maximum (v=V0v = V_0).
The relationship between displacement and velocity is derived from energy conservation:
12mv2+12ky2=12kA2\frac{1}{2} m v^2 + \frac{1}{2} k y^2 = \frac{1}{2} k A^2
where mm is the particle's mass, kk is the wave's spring constant, yy is the displacement, and AA is the amplitude.
Simplifying for Velocity:
From energy conservation:
v2=km(A2−y2)v^2 = \frac{k}{m}(A^2 - y^2)
Substituting V02=kmA2V_0^2 = \frac{k}{m} A^2, we get:
v2=V02(1−y2A2)v^2 = V_0^2 \left(1 - \frac{y^2}{A^2}\right)
When y=A2y = \frac{A}{2}:
v2=V02(1−(A2)2A2)v^2 = V_0^2 \left(1 - \frac{\left(\frac{A}{2}\right)^2}{A^2}\right) v2=V02(1−14)v^2 = V_0^2 \left(1 - \frac{1}{4}\right) v2=V02⋅34v^2 = V_0^2 \cdot \frac{3}{4} v=32V0v = \frac{\sqrt{3}}{2} V_0
Final Answer:
The speed of the particle when its displacement is half the maximum value is:
v=32V0v = \frac{\sqrt{3}}{2} V_0