Question

A wire is attached to a pan of mass $100\text{g}$ that contains a $1\text{kg}$ mass as shown in the figure. When the wire is plucked, it vibrates at a fundamental frequency of $110\text{Hz}$. An additional mass $M$ is then added to the pan and the fundamental frequency changes to $130\text{Hz}$. The value of mass $M$ is

[Assume the pulley to act like a fixed support]

• A. $0.22\text{kg}$
• B. $0.44\text{kg}$
• C. $0.66\text{kg}$
• D. $0.55\text{kg}$

Answer 1

The correct option is B $0.44\text{kg}$

The formula of fundamental frequency is,

$f=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$

where $l$ is length, $T$ is tension and $\mu$ is linear mass density of the wire.

$\Rightarrow$Fundamental frequency $f\propto \sqrt{T}$

$\therefore \frac{f}{f^{\prime }}=\sqrt{\frac{T_1}{T_2}}$

$\Rightarrow \frac{110}{130}=\sqrt{\frac{T_1}{T_2}}$

Initially, mass was $(100\text{g}+1\text{kg})$

So $T_1=(1.1\text{kg}\times g)\text{N}$

When $M$ mass is added,

$T_2=(1.1+M)g\text{N}$

(Where $M$ is in $\text{kg})$

So,

$\frac{110}{130}=\sqrt{\frac{\left(1.1\right)g}{(1.1+M)g}}=\sqrt{\frac{1.1}{1.1+M}}$

$\Rightarrow \frac{(110{)}^2}{(130{)}^2}=\frac{1.1}{1.1+M}$

$\Rightarrow (1.1\times {110}^2)+({110}^2\times M)={130}^2\times 1.1$

$\Rightarrow M=\frac{({130}^2\times 1.1)-(1.1\times {110}^2)}{{110}^2}=0.436\approx 0.44\text{kg}$