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+1 vote
29.8k views
in Physics by (74.3k points)

The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate? 

(a) 1100 Hz

(b) 1000 Hz 

(c) 166 Hz 

(d) 100 Hz

1 Answer

+2 votes
by (70.0k points)
selected by
 
Best answer

Correct option: (b) 1000 Hz

Explanation:

Total length of sonometer wire, l = 110 cm = 1.1 m

Length of wire is in ratio, 6 : 3 : 2 i.e. 60 cm, 30 cm, 20 cm.

Tension in the wire, T = 400 N 

Mass per unit length, m = 0.01 kg

Minimum common frequency = ? 

As we know,

Hence common frequency = 1000 Hz

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