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larisa [96]
4 years ago
10

A guitar string with a linear mass density of 2.0 g/m is stretched between two vertical rods that are 65 cm apart. The string ho

lds a standing wave, fixed at both ends, with three antinodes when it oscillates at a frequency of 430 HzWhat is the frequency of the fifth harmonic of this string?
Physics
1 answer:
Rudiy274 years ago
7 0

Answer:

717 Hz

Explanation:

<u>solution:</u>

The wave with three antinodes has m = 3. Thus, f_3 = 3f_1 and so the fundamental frequency of the string is  

f_1 =f_3/3

    =430 Hz/3

    =143 Hz

Thus, the frequency of the fifth harmonic is

f_5 = 5*f_1

      = 5*143 Hz

     = 717 Hz

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F=\frac{dp}{dt} ............................................(1)

where:

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dt = time taken to change the momentum

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p=m.v

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F=\frac{d(m.v)}{dt}

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\therefore F=m.\frac{dv}{dt}

and, \frac{dv}{dt} =a

where: a = acceleration

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also

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<em>& </em><u><em>Impulse:</em></u>

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For refraction,

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