The vibrations of a string fixed at both ends are described by the equation `y= (5.00 mm) sin [(1.57cm^(-1))x] sin [(314 s^(-1))t]`
(a) What is the maximum displacement of particle at x = 5.66 cm ? (b) What are the wavelengths and the wave speeds of the two transvers waves that combine to give the above vibration ? (c ) What is the velocity of the particle at x = 5.66 cm at time t = 2.00s ? (d) If the length of the string is 10.0 cm, locate the nodes and teh antinodes. How many loops are formed in the vibration ?

Explanation

The amplitude of vibration of a particle at position x is
`A=|5.00 mm sin ([1.57 cm^(-1))x] |` for x =5.66 cm
`A=|(5.00 mm) sin [pi/2xx 5.66]=|(5.00 mm ) sin (2.5 pi + pi/2)]`
`=|(5.00 mm) cos pi/2|` = 2.50 mm.
(b)From the given equation , the wave number `K=1.57cm^(-1)` and angular frequency `omega=314 s^(-1)` , the wavelength is
`lambda-(2pi)/k = (2xx3.14)/1.57 `
`lambda`=4.00 cm and the frequency is `n=omega/(2pi)=314/(2xx3.14)=50 s^(-1)`
Now the wave speed is `v=nlambda=(50 s^(-1))(4.00cm)` = 2.00 m/s
(c )The velocity of the particle at position x at time t is given by
`v=(dy)/(dt)=[(5.00 mm)sin (1.57 cm^(-1))x]314s^(-1) cos (314 s^(-1))t`
`=[(157 cm/s) sin (1.57 cm^(-1)) x ] cos (314 s^(-1))t`
Putting x=5.66 cm and t=2.05 , the velocity of this particle at the given time is
`=(157 cm//s) sin [(5pi)/2+pi/3]cos (200pi)`=(157 cm/s) cos `pi/3 xx1` = 78.5 cm/s
The nodes occur where the amplitude is zero . i.e.
`sin(1.57 cm^(-1))x=0`
`(pi/2 cm^(-1))x=npi`, where n is integer
Thus x=2 n cm
The nodes , occur at x=0 cm 2cm , 4cm , 8 cm , and 10 cm , antinodes occur in between them i.e. at x=1 cm , 3 cm , 5cm , 7 cm and 9 cm . Hence it is clear that string vibrates in five loops .