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
Answer:
Explanation:
frequency of sound waves = 688 Hz
wavelength = 344 / 688 = .5 m
The problem is based on interference of sound waves
For the observer , path difference of sound waves reaching his ear
= 3.5 - 3.00
.5 m
= wavelength
Path difference is equal to wavelength so there will be constructive interference and hence louder sound will be heard by the listener than normal sound as sound waves interfere constructively.
