kmmjones6108
11.02.2020 •
Physics
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
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Ответ:
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
Ответ: