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aleksandr82 [10.1K]
3 years ago
15

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

Physics
1 answer:
nadezda [96]3 years ago
8 0

Answer:

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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