A 660 Hz tuning fork sets up vibration in a string clamped at both ends. The wave speed for a transverse wave on this string is `220 m s^-1` and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion.

Answer

a) 50cm

b) `cos(0.06picm^-1)sin((1320pis^-1)t)`

Explanation

Frequency of the tunning fork,
`f=660Hz`
Wave speed `v=220m/s`
`rarr `lamda`=V/f=220/660=1/3m`
No.of loops = 3
a. So, L.`f=(3/2)v`
`rarr L.660=(3/2)xx220`
` L=1/2m=50cm`
b. the equation of resultant statioN/Ary wave is given by,
`y=2Acos((2pix)/lamda)sin((2pivt)/lamda)`
`rarr y=(0.5)cos((2pix)/(1/3m))sin((2pixx220xxt)/lamda)`
`rarr y=(0.5cm)`
`cos(6pixm^-1)sin(1320pis^-1t)`
`rarr y=(0.5cm)`
`cos(0.06picm^-1)sin(1320pis^-1t)`