JEE Advance - Physics (2009 - Paper 2 Offline - No. 16)
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.
Vastaus
5
Selitys
The distance between the successive nodes is $$\lambda/2$$. Therefore,
$${\lambda \over 2} = {V \over {2v}} = {1 \over {2 \times 100}}\sqrt {{{Tl} \over m}} = {1 \over {200}}\sqrt {{{0.5 \times 0.2} \over {{{10}^{ - 3}}}}} = {1 \over {20}}$$ m = 5 cm
