A sonometer wire supports a 4 kg load and vibrates in fundamental… - Vidyadip

MCQ

A sonometer wire supports a $4\ kg$ load and vibrates in fundamental mode with a tuning fork of frequency $416\ Hz.$ The length of the wire between the bridges is now doubled. In order to maintain fundamental mode, the load should be changed to:

A
$1\ kg$
B
$2\ kg$
C
$8\ kg$
✓
$16\ kg.$

✓

Answer

Correct option: D.

$16\ kg.$

According to the relation of the fundamental frequency of a string $\nu=\frac{1}{2\text{l}}\sqrt{\frac{\text{F}}{\mu}}$
Where $l$ is the length of the string
$F$ is the tension
$\mu$ is the linear mass density of the string
We know that $v_1 = 416\ Hz, l_1 = l$ and $l_2 = 2l$
Also, $m_1 = 4\ kg$ and $m_2 =$ ?
$\nu_1=\frac{1}{2\text{l}_1}\sqrt{\frac{\text{m}_1\text{g}}{\mu}}\ \dots(1)$
$\nu_2=\frac{1}{2\text{l}_2}\sqrt{\frac{\text{m}_2\text{g}}{\mu}}\ \dots(2)$
So, in order to maintain the same fundamental mode
$\nu_1=\nu_2$
squaring both sides of equations $(1)$ and $(2)$ and then equating
$\frac{1}{4\text{l}^2}\frac{4\text{g}}{\mu}=\frac{1}{16\text{l}^2}\frac{\text{m}_2\text{g}}{\mu}$
$\Rightarrow\text{m}_2=16\text{ kg}$

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