Wavelength of traveling wave on string (one end clamped and one end free)

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Homework Help Overview

The discussion revolves around a 30-cm long string that is clamped at one end and free at the other, vibrating in its second harmonic. Participants are trying to determine the wavelength of the traveling waves associated with this setup.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are examining the application of formulas related to harmonic vibrations of strings, questioning the validity of using a formula meant for strings fixed at both ends. There is a focus on the placement of nodes and antinodes in the context of the specific boundary conditions of the string.

Discussion Status

Some participants have provided guidance on correcting the approach to the problem, emphasizing the need to consider the unique characteristics of a string clamped at one end. Multiple interpretations regarding the harmonic number and its implications for the wavelength are being explored.

Contextual Notes

There is a discussion about the constraints of harmonic numbers for strings fixed at one end, noting that only odd harmonics are applicable in this scenario. Participants are also addressing the confusion surrounding the application of formulas for different boundary conditions.

Dalip Saini
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1. Homework Statement
A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wavelength of the constituent traveling waves is
  • A -- 10 cm
  • B -- 40 cm
  • C -- 120 cm
  • D -- 30 cm
  • E -- 60 cm

Homework Equations


for second harmonic n=2
(2L)/n= wavelength

The Attempt at a Solution


(2*30)/(2) = 30 cm
Put the answer is 40 cm, and I don't understand why
 
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Dalip Saini said:

Homework Statement


A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wavelength of the constituent traveling waves is

Homework Equations


for second harmonic n=2
(2L)/n= wavelength

The Attempt at a Solution


(2*30)/(2) = 30 cm
Put the answer is 40 cm, and I don't understand why

The formula you used is wrong. Where are the nodes and antinodes of that vibrating spring?
 
To expand on ehild's answer, you have used the expression for a string where both ends are fixed (or both free). You need to correct forthe fact that your string is fixed in one end and free in the other. This goes for both your threads.
 
Dalip Saini said:
1. Homework Statement
A 30-cm long string, with one end clamped and the other free to move transversely, is vibrating in its second harmonic. The wavelength of the constituent traveling waves is
  • A -- 10 cm
  • B -- 40 cm
  • C -- 120 cm
  • D -- 30 cm
  • E -- 60 cm

Homework Equations


for second harmonic n=2
(2L)/n= wavelength

The Attempt at a Solution


(2*30)/(2) = 30 cm
Put the answer is 40 cm, and I don't understand why
in terms of waves clamped at one end we use the formula 'amplitude=4L/n', however the trick is that their number of harmonics can only be an odd number (1,3,5,7 etc) so the second harmonic must be n=3. by substituting that of the above formula you will get 40 cm. The key words were 'clamp at one end than 2 ends.' for strings clamped at 2 end we use 'amplitude=2L/n' and n can either be an odd or even number.
 

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