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
\[
\text{Acceleration due to gravity}=10\ \text{m s}^{-2}
\]
Show 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 \(T=Mg\).
Step 1: Picture the setup. A string of length $l = 1\ \text{m}$ and total mass $2\ \text{g}$ is fixed at both ends, and a hanging body provides the tension. We want the mass that makes the string ring at its fundamental note of $100\ \text{Hz}$. Step 2: Recall the fundamental frequency. The lowest mode of a string fixed at both ends has frequency \[ f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}, \] where $\mu$ is the linear mass density and $T$ is the tension. Step 3: Get the linear mass density. \[ \mu = \frac{m_{\text{string}}}{l} = \frac{2\times10^{-3}\ \text{kg}}{1\ \text{m}} = 2\times10^{-3}\ \text{kg m}^{-1}. \] Step 4: Solve the frequency relation for $T$. Square the formula to remove the root: \[ f^2 = \frac{T}{4l^2\mu} \quad\Rightarrow\quad T = 4 l^2 \mu f^2. \] Step 5: Plug in the numbers. \[ T = 4(1)^2(2\times10^{-3})(100)^2 = 4 \times 2\times10^{-3}\times 10^4 = 80\ \text{N}. \] Step 6: Convert tension into the hanging mass. The tension equals the weight of the body, so $T = Mg$ with $g = 10\ \text{m s}^{-2}$: \[ M = \frac{T}{g} = \frac{80}{10} = 8\ \text{kg}. \] So a body of $8\ \text{kg}$ is needed. \[ \boxed{8\ \text{kg}} \]