A 20 ~cm long string, having a mass of 1.0 ~g, is fixed at both t… - Vidyadip

MCQ

A $20 \mathrm{~cm}$ long string, having a mass of $1.0 \mathrm{~g}$, is fixed at both the ends. The tension in the string is $0.5 \mathrm{~N}$. The string is set into vibrations using an external vibrator of frequency $100 \mathrm{~Hz}$. Find the separation (in $cm$) between the successive nodes on the string.

✓
$5$
B
$6$
C
$7$
D
$8$

✓

Answer

Correct option: A.

$5$

a
Velocity of wave in a string is $V =\sqrt{ } \frac{\overline{ T }}{\mu}$ where $T =$ tension in the string and $\mu=$ mass per unit length of the string. here, $T =0.5 N$ and $\mu=\frac{1 \times 10^{-3}}{20 \times 10^{-2}}=0.5 \times 10^{-2} \ kg / m$ thus, $v =\sqrt{\frac{0.5}{0.5 \times 10^{-2}}}=10 \ m / s$ Wavelength, $\lambda=\frac{ v }{ f }=\frac{10}{100}=0.1 \ m =10 \ cm$

$\therefore$ The separation between successive nodes $=\frac{\lambda}{2}=\frac{10}{2}=5\ cm$

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