Question 459: Waves

Question

A sonometer wire is in unison with a tuning fork, when it is stretched by weight $w$ and the corresponding resonating length is $L_4$. If the weight is reduced to $\left(\frac{w}{4}\right)$, the corresponding resonating length becomes $L_2$. The ratio $\left(\frac{L_1}{L_2}\right)$ is

Accepted Answer

When sonometer is stretched by weight $w$ (tension), then frequency of vibration in the string
$$\begin{aligned}
& 
u=\frac{1}{2} \sqrt{\frac{T}{m}} \
& 
u=\frac{1}{2 L_1} \sqrt{\frac{w}{m}}\quad	ext{.... (i)}
\end{aligned}$$
Where, $L$ is the resonating length and $m$ is mass per unit length of the string.
Similarly, when weight reduces to $\frac{W}{4}$ and resonating length is $L_2$, then
$$
u=\frac{1}{2 L_2} \sqrt{\frac{\frac{w}{4}}{m}} \Rightarrow v=\frac{1}{4 L_2} \sqrt{\frac{w}{m}}\quad	ext{.... (ii)}$$
$$\begin{aligned}
&	ext { From Eqs. (i) and (ii), we have }\
&\begin{array}{ll} 
& \frac{1}{2 L_1} \sqrt{\frac{w}{m}}=\frac{1}{4 L_2} \sqrt{\frac{w}{m}} \Rightarrow L_1=2 L_2 \
	herefore & \frac{L_1}{L_2}=2 \
	herefore \quad & L_1: L_2=2: 1
\end{array}
\end{aligned}$$

Answer

$4: 1$

Answer

$1: 4$

Answer

$1: 2$

Answer

$2: 1$

Waves

A sonometer wire is in unison with a tuning fork, when it is stretched by weight w and the corresponding resonating length is L_4. If the weight is reduced to \left(\frac{w}{4}\right), the corresponding resonating length becomes L_2. The ratio \left(\frac{L_1}{L_2}\right) is

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