A 20 cm long string, having a mass of 1.9 g, is fixed at both the ends. The tesion in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find te separation (in cm ) between the successive nodes on the string .

Options

A. 3.6 cm
B. 10 cm
C. 15 cm
D. 20 cm

Answer

A

Explanation

We known that , v = `sqrt((T)/(mu))`
where T = tnesion in the string and `mu = ("mass")/("length")`
` v = sqrt( (0.5)/(10^(-3) // 0.2)) = 10 ` m/s
the wavelength of the wave established
`lambda = (v)/(f) = (10)/(100) 0.1 ` m = 10 cm
`therefore` The distance between two successive nodes
= `(lambda)/(2) = (10)/(2) = 5 cm `
So correct choice is (a) .