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.
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Answer

`5`

Explanation

`L=20cm,m=1gm`
`mu=(m)/(L)=(1)/(20)gm//cm=(1)/(20)xx(10^(-3))/(10^(-2))g//m`
`mu=((1)/(200))kg//m,T=0.5N`
`V=sqrt((0.5)/(((1)/(200))))=10m//s,f=100Hz`
`lamda=(V)/(f)=(10)/(100)=(1)/(10)m`
`(lamda)/(2)=(1)/(20)m=5cm`