The length of the wire shown in figure between the pulley is 1.5 m and its mass is 12.0 g. find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.
First let's calculate the tension in the string
Free body diagram of one of the mass
Now for two "loops” to form
we know the string is in its 2nd harmonic
f2 = 2 × v2L = vL
we need to calculate velocity v for that
v = √Tμ and μ= masslength = 12 × 10−31.5
Even though the string has mass ,we can assume tension to be uniform because the mass of the string is
very less compared to the blocks.
v = √90(12 × 10−31.5) = 106.06 ms−1
v2 = 106.061.5 ∼∼ 70 Hz