1.

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

Answer» Correct Answer - 5
`v = sqrt((T)/(mu)) = 10 m//s`
`lamda = (v)/(f) = (10)/(100) = 10 cm`
Distance between the successive nodes `= lamda//2 = 5 cm`.


Discussion

No Comment Found

Related InterviewSolutions