Calculating Time for a Transverse Wave in a Suspended Rope with Mass and Length

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The discussion focuses on calculating the time for a transverse wave to travel the length of a suspended rope with mass and an additional mass at the end. The key equation derived is t = 2√(L / mg) * [√(M+m) - √(M)], which incorporates the effects of tension due to the weight of the mass M and the rope's mass m. The tension in the rope is determined by the weight of the segment below a certain point, leading to a wave speed expression that varies along the rope. Participants emphasize breaking down the problem step by step and understanding the relationship between tension, wave speed, and the mass distribution in the rope. This approach helps clarify the complexities involved in the calculation.
axeeonn
I have this problem due tomorrow, well technically today, and I'm having trouble with it. It askes...

A rope of total mass m and length L is suspended vertically with mass M at the end. Show that the time for a transverse wave to travel the length of the rope is...

t = 2[squ](L / mg)*[[squ](M+m) - [squ](M)]

I start out with t = L/v
where v = [squ](T/[mu]) T being tension and [mu] being mass per unit length

So, t = L/[squ](TL/m)

Thats as far as I got :(

The book hints to find an expression for the wave speed at any point a distance x from the lower end by considering the tension in the rope as resulting from the weight of the segment below that point. But i don't know how to turn that into an equation :(
 
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Find the tension in the rope.

The rope has a mass M at the end of it, that should help you to find the tension.
 


First of all, don't panic! It's completely normal to struggle with a problem like this, especially if it's due soon. Take a deep breath and let's break down the problem step by step.

First, let's review some basic principles. A transverse wave is a type of wave where the disturbance is perpendicular to the direction of propagation. In this case, the rope is being pulled downwards by the weight of the mass M, causing a transverse wave to travel up the rope.

Now, let's look at the equation given: t = 2[squ](L / mg)*[[squ](M+m) - [squ](M)]. This is a complicated equation, but we can break it down into smaller parts to make it easier to understand.

The first part, 2[squ](L / mg), represents the time it takes for the wave to travel the length of the rope. This is similar to the equation you started with, t = L/v, but instead of using the wave speed v, we are using the distance L and the acceleration due to gravity g.

The second part, [[squ](M+m) - [squ](M)], represents the difference in wave speed at the top and bottom of the rope. This is where the hint from the book comes in. By considering the tension in the rope as resulting from the weight of the segment below a certain point, we can find an expression for the wave speed at that point.

Let's break this down further. The tension in the rope at any point x can be represented by T = (M+m)gx/L. This is because the weight of the segment below a certain point is equal to the tension pulling downwards at that point. Now, we can plug this expression for T into the wave speed equation v = [squ](T/[mu]) and get v = [squ]((M+m)gx/[mu]L).

Now, we can plug this expression for v into the original equation for t to get t = 2[squ](L / mg)*[[squ]((M+m)gx/[mu]L) - [squ]((M+m)g/[mu]L)]. This may seem complicated, but remember that [mu] is just the mass per unit length, so [mu]L is just the total mass of the rope.

Now, we can simplify this equation to get t = 2[squ](L / mg)*[[squ
 
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