Question 507: Waves

Question

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.

Accepted Answer

The distance between the successive nodes is $$\lambda/2$$. Therefore,
$${\lambda  \over 2} = {V \over {2v}} = {1 \over {2 	imes 100}}\sqrt {{{Tl} \over m}}  = {1 \over {200}}\sqrt {{{0.5 	imes 0.2} \over {{{10}^{ - 3}}}}}  = {1 \over {20}}$$ m = 5 cm

Waves

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.

Integer / Numeric Answer Type