Solving Wave Propagation & Particle Speed Problems

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The discussion focuses on solving a problem related to wave propagation and particle speed in a string. The equation y(x,t)=A cos[(2(pi)/lambda)(x-vt)] is used to derive the transverse velocity v_y of a particle in the string. To find this velocity, one must analyze the motion of a fixed particle over time, leading to an equation of motion. Once the expression for velocity is established, determining the maximum speed involves basic calculus techniques. The thread emphasizes the importance of understanding the relationship between wave properties and particle dynamics.
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Hi! If anyone could help me with this problem I'd be deeply appreciative:

Speed of Propagation vs. Particle Speed. The equation y(x,t)=A cos2(pi)f (x/v - t) may be written as y(x,t)= A cos[(2(pi)/lamda)(x-vt)]

a) Use the last expression for y(x,t) to find an expression for the transverse velocity v_y of a particle in the string of which the wave travels.

b) Find the maximum speed of a particle of the string?


I really have no idea how I'm even meant to approach this problem... Anyone know how to figure it out?

Thanks!
 
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a) y(x,t) is the height at which the particle of the string at position x and time t is. If you only want to look at one particle of the string, fix x to say x_0 and look only at how the height y(t) varies with time. This, then is the equation of motion of that particle, since it gives its positon as a function of time. How can you find the velocity if you know the equation of motion?

b) One you have found an expression for the velocity in a), finding the maximum value is a basic calculus problem.
 
Io being what? an initial value of 0?
 
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The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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