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3 years ago

Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?

Physics

2 answers:

denpristay [2]3 years ago

3 0

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

$\lambda=\frac{2}{n} L$

The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

$\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m$

The wavelength of the 2nd harmonic is:

$\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m$

The wavelength of the 4th harmonic is:

$\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m$

It is not possible to find any integer n such that $\lambda=5 m$ , therefore the correct options are A, B and D.

kotegsom [21]3 years ago

3 0

For standing waves to be produced,
L = nλ/2, where L is the length of the string, λ is the wavelength and n a natural number.

3.5 = nλ/2
7/λ = n

Therefore, the only answers that produce whole numbers when plugged into λ are A and B.

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Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

$v=\sqrt{\frac{T}{(m/l)}}$

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

$(m/l)=\frac{10}{V^2}$

Explanation:

From the question we are told that

Speed of a transverse wave given by

$v=\sqrt{\frac{T}{(m/l)}}$

Maximum Tension is $T=10.0N$

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$v=\sqrt{\frac{T}{(m/l)}}$

$v^2=\frac{T}{(m/l)}$

$(m/l)=\frac{T}{V^2}$

$(m/l)=\frac{10}{V^2}$

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$(m/l)=\frac{10}{V^2}$

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Answer:

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Explanation:

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