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A stretched string is 120 cm long and has a linear density of 0.022 g/cm. What tension in the string will result in a second harmonic of 590 Hz

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Final answer:

To find the tension that results in a second harmonic of 590 Hz for a string with a given length and linear mass density, we use the formula f = (n/2L) * sqrt(T/μ). Solving for T gives us a tension of 146.21 N.

Step-by-step explanation:

To determine what tension in the string will result in a second harmonic of 590 Hz, we can use the formula for the frequency of a standing wave on a string, which is given by:

f = (n/2L) * sqrt(T/μ)

Where f is the frequency, n is the harmonic number, L is the length of the string, T is the tension in the string, and μ is the linear mass density of the string.

Given:

μ = 0.022 g/cm = 0.022 kg/m (the linear mass density in kg/m)

Solving for T, we have:

T = (2Lf)^2 μ = (2 * 1.20 m * 590 Hz)^2 * 0.022 kg/m

T = 146.21 N (rounded to two decimal places)

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