Question

An object of specific gravity \(\rho\) is hung from a thin steel wire. The fundamental frequency for the transverse stationary waves is 300 Hz. The object is immersed in water, so that one half of its volume is submerged. The second harmonic frequency will be ( density of water \(=1\) g / cc)

• A. \(600\left(\frac{2\rho -1}{2\rho }\right)^{1/2}\)
• B. \(600\left(\frac{2\rho }{2\rho -1}\right)^{1/2}\)
• C. \(600\left(\frac{\rho -1}{2\rho }\right)^{1/2}\)
• D. \(600\left(\frac{\rho }{2\rho -1}\right)^{1/2}\)

Hint:

The wire tension equals the apparent weight, which falls when half of the object is submerged, and frequency is proportional to the square root of tension.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The fundamental frequency of a wire is \(f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}\), so \(f\propto\sqrt{T}\). The tension equals the apparent weight of the object.

Step 2: Calculation
In air: \(T_{1}=V\rho g\) gives \(f_{1}=300\ \text{Hz}\). Half immersed: upthrust \(=\frac{V}{2}\times1\times g\), so \(T_{2}=V g\left(\rho-\frac{1}{2}\right)\).

Step 3: Frequency
\(f_{2}=300\sqrt{\frac{\rho-1/2}{\rho}}=300\sqrt{\frac{2\rho-1}{2\rho}}\). The second harmonic is twice this: \(600\left(\frac{2\rho-1}{2\rho}\right)^{1/2}\).

Final Answer:
The second harmonic frequency is 600 times the square root of (2 rho minus 1) over 2 rho. \[ \boxed{\text{(A) }600\left(\dfrac{2\rho-1}{2\rho}\right)^{1/2}} \]