Question

A body is suspended from a string of length \(1\ \text{m}\) and mass \(2\ \text{g}\). The mass of the body to produce a fundamental mode of \(100\ \text{Hz}\) frequency in the string is
\[ \text{Acceleration due to gravity}=10\ \text{m s}^{-2} \]

• A. \(80\ \text{g}\)
• B. \(4\ \text{kg}\)
• C. \(400\ \text{g}\)
• D. \(8\ \text{kg}\)

Hint:

For the fundamental mode of a stretched string, use \(f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\), where tension due to a hanging mass is \(T=Mg\).

Answer 1

The Correct Option is D

Step 1: Write the formula for fundamental frequency.
For a stretched string, the fundamental frequency is
\[ f=\frac{1}{2l}\sqrt{\frac{T}{\mu}} \]
where \(l\) is the length of the string, \(T\) is the tension, and \(\mu\) is the mass per unit length.

Step 2: Find mass per unit length.
Given mass of string is
\[ 2\ \text{g}=2\times10^{-3}\ \text{kg} \]
Length of string is
\[ l=1\ \text{m} \]
So,
\[ \mu=\frac{2\times10^{-3}}{1} \]
\[ \mu=2\times10^{-3}\ \text{kg m}^{-1} \]

Step 3: Find the tension in the string.
Given,
\[ f=100\ \text{Hz} \]
Using
\[ f=\frac{1}{2l}\sqrt{\frac{T}{\mu}} \]
\[ 100=\frac{1}{2(1)}\sqrt{\frac{T}{2\times10^{-3}}} \]
\[ 200=\sqrt{\frac{T}{2\times10^{-3}}} \]
Squaring both sides,
\[ 40000=\frac{T}{2\times10^{-3}} \]
\[ T=40000\times2\times10^{-3} \]
\[ T=80\ \text{N} \]

Step 4: Find the suspended mass.
The tension is produced by the weight of the suspended body.
So,
\[ T=Mg \]
\[ 80=M(10) \]
\[ M=8\ \text{kg} \]

Step 5: Final conclusion.
Hence, the mass of the body is
\[ \boxed{8\ \text{kg}} \]