How to Calculate Time, Weight, and Wavelength in a String with Weight System?

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To calculate the time it takes for a pulse to travel the length of a 1.60 m string with a wave speed of 15.9 m/s, the time is found to be approximately 0.101 seconds. The weight W supported by the string is determined to be 21 N, which corresponds to the tension in the string. The wavelength of the wave is calculated as 0.0365 m. To find how many wavelengths fit on the string, the total length of the string is divided by the wavelength, resulting in approximately 43.8 wavelengths. The discussion emphasizes understanding wave properties and the relationship between tension, weight, and wave speed in a string system.
Vanessa Avila
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Homework Statement


A 1.60-m string of weight 1.30 N is tied to the ceiling at its upper end, and the lower end supports a weight W. When you pluck the string slightly, the waves traveling up the string obey the equation y(x,t)=(8.50mm)cos(172rad⋅m−1(x)−2730rad⋅s−1(t))

a) How much time does it take a pulse to travel the full length of the string?
b) What is the weight W?
c) How many wavelengths are on the string at any instant of time?

Homework Equations


Acos2π(x/λ- t/T)
k = 2π/λ
w = 2π/T

v =√(F/μ)

The Attempt at a Solution


I found the mass of the string by using F = mg and got 0.133 kg. With this mass, I was able to find the density (m/L) which is 0.0831m/kg

for the time, I thought i should do
2730rad/s = 2π/T
T = 2π/2730 rad/s

for how many wavelengths, I thought of doing:
172rad/m = 2π/λ
λ = 2π/172 rad/m, but that was wrong.
 
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Vanessa Avila said:
I found the mass of the string by using F = mg and got 0.133 kg. With this mass, I was able to find the density (m/L) which is 0.0831m/kg
OK, except for the units.

for the time, I thought i should do
2730rad/s = 2π/T
T = 2π/2730 rad/s
You calculated a time here. But is it the time that is asked for in part (b)? What does the symbol T stand for in your calculation?

for how many wavelengths, I thought of doing:
172rad/m = 2π/λ
λ = 2π/172 rad/m, but that was wrong.
I think you calculated λ correctly. But part (c) is asking for "how many wavelengths are on the string".
 
TSny said:
OK, except for the units.

You calculated a time here. But is it the time that is asked for in part (b)? What does the symbol T stand for in your calculation?

I think you calculated λ correctly. But part (c) is asking for "how many wavelengths are on the string".
Oh T stands for the period. How do we use the period to find small t? and how do we find how many wavelengths are on the string? and my bad! I mixed up the units for density. It should be kg/m. Thanks for that!
 
Can you find the speed of a wave pulse using the information given? Hint: the speed of a wave pulse is the same as the wave speed of the wave y(x,t) that is given in the problem.

I think "how many wavelengths are on the string" means how many wavelengths fit into the length of the string.
 
TSny said:
Can you find the speed of a wave pulse using the information given? Hint: the speed of a wave pulse is the same as the wave speed of the wave y(x,t) that is given in the problem.

I think "how many wavelengths are on the string" means how many wavelengths fit into the length of the string.
So it's v = λ/T ?

T = 2.302E-3
λ = 0.0365 m

v = (0.0365/2.302E-3) = 15.9 m/s
 
Your answer for v looks right.
 
TSny said:
Your answer for v looks right.
Would I then solve for time by doing v = distance/time by using the length of the string as distance?
15.9 m/s = 1.60m/t
t = 1.60 /15.9 = 0.101s
 
Vanessa Avila said:
Would I then solve for time by doing v = distance/time by using the length of the string as distance?
15.9 m/s = 1.60m/t
t = 1.60 /15.9 = 0.101s
OK
 
TSny said:
OK
:) My answer for t was correct for that one. yay!

And now for the weight, I'm not sure where to start.
 
  • #10
The weight W probably has something to do with the tension in the string.
 
  • #11
TSny said:
The weight W probably has something to do with the tension in the string.
would it be 1.30 N as well since it's pulling the weight up?
 
  • #12
Consider a free body diagram for the object hanging on the end of the string.
 
  • #13
TSny said:
Consider a free body diagram for the object hanging on the end of the string.
Oh woop. It has Tension up and mg down. I don't have T, but i can solve that by v = √(Ft/μ)

So my Ft is 21 N which means W is also 21 N. That answr was correct! :)

Now for finding how many wavelengths, do I use the λ I solved?
 
  • #14
Vanessa Avila said:
Oh woop. It has Tension up and mg down. I don't have T, but i can solve that by v = √(Ft/μ)

So my Ft is 21 N which means W is also 21 N. That answr was correct! :)
Great!

Now for finding how many wavelengths, do I use the λ I solved?
Sure.
 
  • #15
TSny said:
Great!

Sure.
But what equation do I use to solve for the n numbers of wavelenghts?
 
  • #16
Suppose the wave y(x, t) that is given in the problem exists on the string. If you took a picture of the string at a particular instant of time, how many wavelengths would you count along the string?
 
  • #17
The problem is a little confusing. If you "pluck" the string to get a "pulse", it would not have the form y(x, t) that is given in the problem (which represents a "sinusoidal wave" spread all along the string).
 
  • #18
TSny said:
The problem is a little confusing. If you "pluck" the string to get a "pulse", it would not have the form y(x, t) that is given in the problem (which represents a "sinusoidal wave" spread all along the string).
So if it's just a pulse then it's just 1?
 
  • #19
TSny said:
The problem is a little confusing. If you "pluck" the string to get a "pulse", it would not have the form y(x, t) that is given in the problem (which represents a "sinusoidal wave" spread all along the string).
oh wait. So i have a string 1.60 m long, and each wavelength is 0.0365
should it then be
1.60m/0.0365 to find how many there are?
 
  • #20
Vanessa Avila said:
So if it's just a pulse then it's just 1?
But, the function y(x, t) given in the problem is not really a pulse. It is a sinusoidal wave spread along the string. So, there are a certain number of wavelengths that would occupy the string at a given instant of time.
 
  • #21
Vanessa Avila said:
oh wait. So i have a string 1.60 m long, and each wavelength is 0.0365
should it then be
1.60m/0.0365 to find how many there are?
Yes!
 
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  • #22
TSny said:
Yes!
alright! Awesome! Thank you so much :)
 
  • #23
You are welcome. Good work.
 
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