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

AI Thread Summary
The discussion centers on calculating the wavelength of a string vibrating in its second harmonic, with one end clamped and the other free. The correct formula for this scenario is based on the fact that only odd harmonics are present, leading to the conclusion that the second harmonic corresponds to n=3. Thus, the wavelength is calculated using the formula (4L/n), resulting in a wavelength of 40 cm for a 30-cm string. The initial misunderstanding arose from incorrectly applying the formula for strings fixed at both ends. Understanding the boundary conditions is crucial for determining the correct wavelength in such systems.
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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