question_answer

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?                                           [JEE ONLINE 19-04-2014]

A)  1100 Hz                      

B)  1000 Hz

C)  166 Hz                        

D)  100 Hz

Correct Answer: B

Solution :

[b] Total length of sonometer wire,

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,

Frequency,

Similarly,

Hence common frequency = 1000 Hz