Question

006 10.0 points A stretched string is 200 cm long and has a linear density of 0.012 g/cm. What tension in the string will result in a second harmonic of 490 Hz? Answer in units of N.

Answer 1

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Given that,

Length of the string,

L = 200 cm = 2 m

Linear density of string,

u=0.012 g/cm = 0.012 x 10-3 kg/10-2 m = 0.0012 kg/m

Second harmonic frequency,

f = 490 H2

For, second harmonic, frequency is given by:

T

Where T is the tension in the string.

me/9 Ekov o x 147 -- ook

Therefore, Tension in the string is given by:

T = (490 Hz x 2 m) x 0.0012 kg/m

::T= 1152.48 N

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