A wire of the length 0.5 m is stretched by a weight of 2 kg. If the mass per unit length of the wire is 1.96 ×
10-3
kg/m, find the fundamental frequency of the wire and frequency of its first overtone.
Answers
The fundamental frequency of wire is 100 Hz
Frequency of first overtone is 200 Hz
Explanation:
Length of the wire m
Weight attached kg
Therefore, tension in the wire
N
Mass per unit length kg/m
We know that fundamental frequency of vibration is given by
s⁻¹
Hz
Frequency of first overtone
Hz
Hope this answer is helpful.
Know More:
Q: Illustrate using labelled diagrams only, a sonometer wire of length, l, vibrating at its(i) fundamental(ii) first overtone(iii) second overtone.
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Q: A sonometer wire of length 0.5 m is stretched by a weight of 5 kg. The fundamental frequency of vibration is 100Hz. Determine the linear density of material of wire. (Ans : 0.0049 kg/m)
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Given :
The length of wire , l = 0.5 m
Weight = 2 kg
Mass per unit , μ = 1.96 x 10 ^ (-3)
To find :
Fundamental frequency and frequency of the first overtone .
Solution :
Tension , T = mass * g
= 2 * 9.8
= 19.6 N
fundamental frequency , v = ( 1 / 2*l ) * √( T / μ )
= ( 1 / 1 ) * √( 19.6 / 1.96 x 10 ^ (-3) )
= √10000
= 100 Hz
frequency of first overtone = 2 * fundamental frequency
= 2 * 100 Hz
= 200 Hz
The fundamental frequency is 100Hz and the frequency of the first overtone is 200 Hz .