A string of length 1 m and mass 5 g is fixed at both ends. The… | JEE Main 2019 (10 Jan, Shift 1)

Q94JEE Main 2019 · 10 JAN · SHIFT 1Medium

Stationary Waves & Vibrating Strings

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to -

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Answer: (c)

$\mu = \dfrac{5\times10^{-3}}{1} = 5\times10^{-3}\text{ kg/m}$ , $v = \sqrt{\dfrac{8}{5\times10^{-3}}} = 40\text{ m/s}$ .
The string vibrates at the driver frequency: $\lambda = \dfrac{v}{n} = \dfrac{40}{100} = 0.4\text{ m}$ .
Successive nodes are $\dfrac{\lambda}{2} = 0.2\text{ m} = 20\text{ cm}$ apart.
So the answer is (c) 20.0 cm.

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