Time needed for a transverse wave to propagate on a string

In summary, the conversation discusses finding the time t in terms of L, mu, and T_s for a string of total length L, made up of three segments of equal length with varying mass per unit length. The time is expressed as ((1+(1/sqrt(2))+2)*sqrt(T_s/mu))/(L/3) and can also be found by adding the individual times for each segment (t_1, t_2, t_3).
  • #1
spaghed87
24
1

Homework Statement


Consider a string of total length L, made up of three segments of equal length. The mass per unit length of the first segment is mu, that of the second is 2*mu, and that of the third mu/4. The third segment is tied to a wall, and the string is stretched by a force of magnitude T_s applied to the first segment; T_s is much greater than the total weight of the string.

Express the time t in terms of L, mu, and T_s. I must use those variables for this answer, no values were given for those variables so they must be included in the answer.


Homework Equations



Velocity=sqrt(T_s/mu) -velocity of string

where, T_s is the tension of the string and mu is its linear denisty. mu=mass/length


The Attempt at a Solution



My answer

time=((1+(1/sqrt(2))+2)*sqrt(T_s/mu))/(L/3)

since, time=v/m ==> m/s/m = seconds right? Edit: = 1/seconds

How I got that:

By pulling out the number multiplied by mu from the sqaure root I get:

The first segment is a velocity of 1*sqrt(T_s/mu)

The second segment is a velocity of 1/sqrt(2)*sqrt(T_s/mu)

The third segment is a velocity of 2*sqrt(T_s/mu)

Add those velocities toegther. Then to get the time you can just divide the velocity by the length of the string that is divided into three equal segments. So, dividing by (L/3) will give the answer. Anyone see what is wrong with my answer?
 
Last edited:
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  • #2
spaghed87 said:
... time=v/m ==> m/s/m = seconds right?

Well, no, those units would be 1/seconds or seconds-1, so something is wrong.

Your equation says that the faster the wave, the longer it would take, which is wrong intuitively. Faster waves should take less time to travel along the string.
 
  • #3
I was thinking I had 1/s for time instead of s but I was having a brain fart. So, if I take the inverse of that equation it should be right then?
 
  • #4
I ended up solving it differently by finding t_1, t_2, t_3, and then by adding all of the times for the time overall. Woot extra credit for me. Thanks for the help. Time for the test tomorrow. :smile:
 

Related to Time needed for a transverse wave to propagate on a string

1. How is the time needed for a transverse wave to propagate on a string calculated?

The time needed for a transverse wave to propagate on a string is calculated using the formula T = L/v, where T is the time in seconds, L is the length of the string in meters, and v is the velocity of the wave in meters per second.

2. What factors affect the time needed for a transverse wave to propagate on a string?

The time needed for a transverse wave to propagate on a string is affected by the length of the string, the tension of the string, and the density of the string. A longer string, higher tension, and lower density will result in a shorter propagation time.

3. Does the amplitude of the wave affect the time needed for propagation?

No, the amplitude of the wave does not affect the time needed for propagation. The time needed is solely determined by the length, tension, and density of the string.

4. Can the time needed for a transverse wave to propagate on a string be changed?

Yes, the time needed for a transverse wave to propagate on a string can be changed by altering the length, tension, or density of the string. Additionally, changing the medium through which the wave is traveling can also impact the propagation time.

5. Why is the time needed for a transverse wave to propagate on a string important?

The time needed for a transverse wave to propagate on a string is important because it helps us understand the behavior and characteristics of waves. It also plays a crucial role in various applications, such as musical instruments, telecommunications, and earthquake detection.

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