Tranverse velocity of a point on a string

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The discussion focuses on calculating the transverse velocity of a point on a string described by the sinusoidal wave equation y(x,t) = 0.12sin[8 π(t - x/50)]. The key point is that transverse velocity refers to the movement of a point on the string in the y-direction, not along the string itself. The correct approach involves taking the derivative of the displacement with respect to time, resulting in dy/dt = 0.12 * 8π * cos(8πt - 8πx/50). After substituting the values for x and t, one participant initially calculated the transverse velocity as 1.6 m/s but later corrected it to 2.868 m/s. The discussion highlights the importance of understanding the distinction between transverse velocity and wave speed.
jegues
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Homework Statement



A sinusoidal wave is moving along a string. The equation governing the displacement as a function of position and time is,

y(x,t) = 0.12sin[8 \pi(t-\frac{x}{50})],

where x and y are in meters, and t is in seconds. At t = 2.4s, what is the transverse velocity of a point on the string at x = 5.0m?

Homework Equations





The Attempt at a Solution



I don't know how to get started on this one.
 
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jegues said:
I don't know how to get started on this one.
"Transverse" means perpendicular to the string. The "transverse velocity" is not the speed of the wave. Rather it is the velocity of a tiny point on the string itself (attached to the string).

You are given the displacement of that point on a string, using the given equation,

<br /> y(x,t) = 0.12sin[8 \pi(t-\frac{x}{50})],<br />

What's the relationship between displacement and velocity (in terms of integrals, derivatives, etc.)?
 
collinsmark said:
"Transverse" means perpendicular to the string. The "transverse velocity" is not the speed of the wave. Rather it is the velocity of a tiny point on the string itself (attached to the string).

You are given the displacement of that point on a string, using the given equation,

<br /> y(x,t) = 0.12sin[8 \pi(t-\frac{x}{50})],<br />

What's the relationship between displacement and velocity (in terms of integrals, derivatives, etc.)?

Velocity is just \frac{dx}{dt} isn't it?
 
jegues said:
Velocity is just \frac{dx}{dt} isn't it?
dx/dt is the change in position per unit time (i.e. velocity) of something along the length of the string, assuming the string lies along the x-axis.

But a point on the string itself does not move along length of the string. It moves in a perpendicular, transverse direction. Specifically, it moves in the y direction. :wink: You're looking for dy/dt.
 
collinsmark said:
dx/dt is the change in position per unit time (i.e. velocity) of something along the length of the string, assuming the string lies along the x-axis.

But a point on the string itself does not move along length of the string. It moves in a perpendicular, transverse direction. Specifically, it moves in the y direction. :wink: You're looking for dy/dt.

Okay so,

\frac{dy}{dt} = 0.12 \cdot 8\pi cos(8 \pi t - \frac{8 \pi x}{50})

When I plug the numbers in I get,

\frac{dy}{dt} = 1.6m/s

Which is still incorrect?
 
Bump, still looking for help on finishing this one off!
 
Check the evaluation, if you did not mix radians with degrees.

ehild
 
jegues said:
Bump, still looking for help on finishing this one off!
I got 2.868 m/s.
 

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