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

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