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

Waves PYQ — JEE Advanced 2009 · Shift 2 · Q54 with Solution

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Correct answer: 5

The linear mass density is

\mu = \frac{m}{L} = \frac{1.0 \times 10^{-3}}{0.20} = 5.0 \times 10^{-3}\text{ kg/m}

For a stretched string,

v = \sqrt{\frac{T}{\mu}}

Substituting the values,

v = \sqrt{\frac{0.5}{5.0 \times 10^{-3}}} = \sqrt{100} = 10\text{ m/s}

Using

v = f\lambda

we get

\lambda = \frac{v}{f} = \frac{10}{100} = 0.10\text{ m} = 10\text{ cm}

In a stationary wave, the distance between two successive nodes is

\frac{\lambda}{2}

So,

\frac{\lambda}{2} = \frac{10\text{ cm}}{2} = 5\text{ cm}

The separation between successive nodes is

\boxed{5\text{ cm}}