Physics, 04.01.2021 16:20 talia43

Standing waves are created in the four strings shown in Figure 25. All strings have the same mass per unit length and are under the same tension The lengths of the strings are given. Rank the frequencies of the oscillations, from largest to smallest

Standing waves are created in the four strings shown in Figure 25. All strings have the same mass p

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Answer from: kaylaamberd

The order of the frequencies is f_Df_Bf_Af_C

Standing waves:

The frequency of the standing wave in a string tied at both ends is given by:

$f=\frac{nv}{2L}$

where n is the mode of frequency

v is the velocity of the wave

and L is the length of the string.

Now the velocity of a wave in a string tied at both ends is given by

$v=\sqrt{\frac{T}{\mu}}$

where T is the tension and μ is the mass per unit length.

Since T and μ are the same for all the strings, velocity will be the same for all.

Now to find the mode of frequency we can calculate the number of nodes (including the nodes at the ends) in the given figure and subtract by 1. Nodes are the point where the amplitude of the wave is zero.

$f_A=\frac{3v}{2\times27}=\frac{v}{18}\;s^{-1}\\\\f_B=\frac{4v}{2\times30}=\frac{v}{15}\;s^{-1}\\\\f_C=\frac{3v}{2\times30}=\frac{v}{20}\;s^{-1}\\\\f_D=\frac{4v}{2\times24}= \frac{v}{12}\;s^{-1}$

Hence, f_Df_Bf_Af_C

Learn more about standing waves:

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Answer from: ghostboy2001

The rank of the frequencies from largest to smallest is

The largest frequency of oscillation is given by the string in option D

The second largest frequency of oscillation is given by the string in option B

The third largest frequency of oscillation is given by the string in option A

The smallest frequency of oscillation is given by the string in option C

Explanation:

The given parameters are;

The mass per unit length of all string, m/L = Constant

The tension of all the string, T = Constant

The frequency of oscillation, f, of a string is given as follows;

$f = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L}$

Where;

T = The tension in the string

m = The mass of the string

L = The length of the string

n = The number of overtones

$Therefore, \ {\sqrt{\dfrac{T}{m/L} } } = Constant \ for \ all \ strings = K$

For the string in option A, the length, L = 27 cm, n = 3 we have;

$f_A = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 27} = \dfrac{2 \times K}{27} \approx 0.07407 \cdot K$

For the string in option B, the length, L = 30 cm, n = 4 we have;

$f_B = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 30} = \dfrac{ K}{12} \approx 0.08 \overline 3\cdot K$

For the string in option C, the length, L = 30 cm, n = 3 we have;

$f_C = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 30} = \dfrac{K}{15} \approx 0.0 \overline 6 \cdot K$

For the string in option D, the length, L = 24 cm, n = 4 we have;

$f_D = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 24} = \dfrac{5 \times K}{48} \approx 0.1041 \overline 6 \cdot K$

Therefore, we have the rank of the frequency of oscillations of th strings from largest to smallest given as follows;

1 ) f_D 2) f_B 3) f_A 4) f_C

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