The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in the 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?

Answer is B
Since the length of sonometer is divided in the ratio 6:3:2, then the lengths of the three segments are:
Length of segment 1 = (6/11) * 110 cm = 60 cm = 0.6 m
Length of segment 2 = (3/11) * 110 cm = 30 cm = 0.3 m
Length of segment 3 = (2/11) * 110 cm = 20 cm = 0.2 m
The maximum wavelength for each segment is twice its length:
λ1 = 2 * 0.6 m = 1.2 m
λ2 = 2 * 0.3 m = 0.6 m
λ3 = 2 * 0.2 m = 0.4 m
The speed of waves on a string is given by:
V = √(T/μ)
where T is the tension (400 N) and μ is the mass per unit length (0.01 kg/m).
V = √(400 N / 0.01 kg/m) = √(40000 m²/s²) = 200 m/s
The frequency for each segment is:
f1 = V/λ1 = 200 m/s / 1.2 m = 166.67 Hz
f2 = V/λ2 = 200 m/s / 0.6 m = 333.33 Hz
f3 = V/λ3 = 200 m/s / 0.4 m = 500 Hz
To find the minimum common frequency, we need to find the least common multiple (LCM) of f1, f2, and f3.
The prime factorization of these frequencies are approximately:
f1 ≈ 166.67 ≈ 2 * 83.33
f2 ≈ 333.33 ≈ 3 * 111.11
f3 = 500 = 2² * 5³
Finding the LCM is not straightforward with these approximate values. However, let's reconsider the wavelength calculation. The maximum wavelength is not necessarily twice the length of each segment, as the question asks for the minimum common frequency. If we assume that each segment vibrates at its fundamental frequency, then the wavelength is twice the segment length. Then:
f1 = V / (2 * 0.6) = 200/1.2 = 166.67 Hz
f2 = V / (2 * 0.3) = 200/0.6 = 333.33 Hz
f3 = V / (2 * 0.2) = 200/0.4 = 500 Hz
The least common multiple of these frequencies is difficult to determine precisely from these approximate values. However, a closer examination reveals that these frequencies are approximately multiples of 100Hz: (166.67, 333.33, 500). This suggests that 100Hz is a strong candidate for a common harmonic. Let's re-examine the problem using the fundamental frequency. The fundamental frequency is given by:
f = V / (2L)
where L is the length of the segment. Therefore the fundamental frequencies of the three segments are:
f1 = 200/(20.6) = 166.67 Hz
f2 = 200/(20.3) = 333.33 Hz
f3 = 200/(2*0.2) = 500 Hz
The minimum common frequency is 100 Hz.