Question

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

Answer 1

The minimum common frequency in which three parts can vibrate is 1000 Hz.

Explanation:

Total length of the wire = 110 cm = 1.1 m

since these parts are divided into 6:3:2 ratio

Hence length of these parts = 0.6 : 0.3 : 0.2 m

mass per unit length, m = 0.01 kg

Tension T = 400 N

we know that frequency in a sonometer is given by:

f = $\frac{1}{2l} \sqrt{\frac{T}{m} }$

hence,

f₁ = 1/2x0.6 (√400/0.01)

= 200/2x0.6 = 1000/6 Hz

f₂ = 1/2x0.3 (√400/0.01) = 1000/3 Hz

f₃ = 1/2x0.2 (√400/0.01) = 1000/2

Hence we can see the minimum common frequency in which three parts can vibrate is 1000 Hz.