String with a linear mass density of \[\mu =0.0062\,\text{kg/m}\] is stretched between two posts 1.30 m apart. The tension in the string is 150.00 N. The string oscillates and produces a sound wave. A 1024-Hz tuning fork is struck and the beat frequency between the two sources is 52.83 Hz. What are the possible frequency and wavelength of the wave on the string?

A) The possible values of frequency of the wave of the string are 971.17 Hz or 1076.83 Hz

B)The possible values of wavelength of the wave of the string are 0.14m or 0.16m

Explanation:

We are given;

Linear mass density of string; μ = 0.0062 kg/m

Length of string; L = 0.3m

Tension in string; T = 150 N

Frequency of tuning fork; f_1 = 1024 Hz

Frequency of the beat; f_b = 52.83 Hz

Now, we know that beat frequency is given by the formula;

f_b = f_1 - f_2 or f_b = f_2 - f_1

Where f_2 is the wave of the string.

Thus;we can calculate the 2 possible values of f_2

52.83 = 1024 - f_2

f_2 = 1024 - 52.83 = 971.17 Hz

Or;

52.83 = f_2 - 1024

f_2 = 1024 + 52.83 = 1076.83 Hz

So, f_2 is either 971.17 Hz or 1076.83 Hz

Now, the wavelength is given by the formula;

λ = (1/f)√(T/μ)

Now, applying it to the wavelength of the string, f will be f_2

So wavelength will have 2 possible values;

At f_2 = 971.17 Hz

λ = (1/971.17)√(150/0.0062) = 0.16m

At f_2 = 1076.83 Hz

λ = (1/1076.83)√(150/0.0062) = 0.14m

The possible values of wavelength are 0.14m or 0.16m

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your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

see earlier answer