Question

When mass $M$is tied to the bottom of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is$200\mathrm{Hz}$Adding an additional$1.0\mathrm{kg}$to the hanging mass increases the second-harmonic frequency$245\mathrm{Hz}$. What is $M$?

Short answer

The suspended mass is $2$kg.

Step-by-step solution

Given information

Second harmonic frequency when mass$M$is suspended$f=200\mathrm{Hz}.$

When is suspended, $f^{\text{'}}=245\mathrm{Hz}$(second harmonic frequency)

Length of wire $\mathrm{L}=$(same for both frequencies as the same wire is used for both masses)$\mu =$(Same for both frequencies as the same wire is used)

Formula:

$\mathrm{f}=\frac{1}{L}\sqrt{\frac{T}{\mu }}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..$

(General frequency for the second harmonic)

$\mathrm{T}=Mass\mathrm{x}gravity.........................$

(Tension in the wire due to suspended masses)

Calculation

(I) When the mass $\text{'}M\text{'}$is suspended

$$\begin{gathered} f=\frac{1}{L}\sqrt{\frac{T}{\mu }} \\ 200=\frac{1}{L}\sqrt{\frac{M_g}{\mu }} \end{gathered}$$

Both sides are squaring up

$40000=\frac{1}{L^2}\frac{M_R}{\mu }$

(II) when the mass $\text{'}M+1\text{'}$is suspended.

${\mathrm{f}}^{\text{'}}=\frac{1}{L}\sqrt{\frac{T}{\mu }}$

$245=\frac{1}{L}\sqrt{\frac{(M+I)}{\mu }}$

Both sides are squaring up

$60025=\frac{1}{L^2}\frac{(M+1)g}{\mu }$

Dividing equation (I) from equation (II)

We compute that the initial mass hanging from the wire is $2$kg using both frequencies.