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A string is stretched to a length of 308 cm and both ends are fixed. If the density of the string is 0.023 g/cm, and its tension is 967 N, what is the fundamental frequency? Answer in units of Hz

Respuesta :

Answer:

Frquency=3,994Hz

Explanation:

Tension =967N

Density of string (μ)=0.023g/cm

Length of the stretched spring=308cm

Fundamental frequency for nth harmonic :

Fn=n/2L(√T/μ)

Substituting the given values to find the frequency :

f1=1/2(308cm) *(0.01m/1cm)[(√967N)/(0.023g/cm)(0.1kg)/(0.1kg/m)/(1g/cm)]

=6.16m[(√967N)/0.0023kg/m)]

=3,994.20Hz

Approximately,

The frequency will be =3,994Hz

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