# A metal wire of diameter 1 mm is held on two knife edges separated by a distance of 50 cm. The tension in the wire is 100 N. The wires, vibrating with its fundamental frequency and a vibrating tuning fork together produce 5 beats/ sec. The tension in the wire is [hen reduced to 81 N. When the two are excited, beats are created again at the same rate. Calculate.(i)  The frequency of the fork.(ii) The density of the material of the wire.

10 years ago
Hello Student,
V = 1/2ℓ √T/m = 1/2 x 0.5 √100/m = 10√m . . . . . . . . . . . . . . . . . . . . . . . . . (i)
The frequency of the tuning fork is either v + 5 or v – 5.
NOTE : On decreasing the tension, the frequency will decrease.
Therefore the frequency of tuning fork should ve v – 5.
Now, v1 = 1/2ℓ √T/m = 1/2 x 0.5 √81/m = 9/√m
This v1 should be v – 10.
∴ 10/√m – 10 = 9/√m ⇒ 10 – 9/√m = 10
M = 1/100 . . . . . . . . . . . . . . . . . . (ii)
P x π d2/4 = 1/100 (m =density x volume)
⇒ p = 4/100 x π x (10-3 )2 = 12738.85 kg/m3
From (i) and (ii) v = 10/√1/100 = 100 Hz
∴ Frequency of the fork is 95 Hz

Thanks