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

(a) The amplitude of the vibration of the particle at position x is
`A | (5.00 mm) sin[(1.57cm^(-1))x]|`
For x = 5.66 cm
`A =|(5.00mm) sin[(pi)/(2)xx5.66]|`
`=|(5.00 mm)sin(2.5pi + (pi)/(3))|`
`=| (5.00mm)cos (pi)/(3) | = 2.50mm.`
(b) From the given equation the wave number `k = 1.57cm^(-1)` and the angular frequency `omega = 314s^(-1)` thus, the wavelength is
`lambda= (2pi)/(k) = (2xx3.14)/(1.57cm^(-1)) = 4.00cm`
and the frequancy is `v = (omega)/(2pi) = (314s^(-1))/(2xx3.14) = 50s^(-1).`
The wave speed is `upsilon = vlambda = (50 s^(-1) (4.00cm) = 2.00 m s^(-1).` (c ) The velocity of the particle of the particla positiion x =at time t is given by `upsilon = (dely)/(delt) = (5.00 mm) sin[(1.57cm^(-1))x]`
`[314 s^(-1) cos (314 s^(-1)t]`
`=(157 cm s^(-1)) sin (1.57 cm^(-1)) x cos (314 s^(-1))t.`
Putting x = 5.66 cm and t = 2.00 s, the velocity of this particvle at the given instant is
`(157 cm s^(-1) sin ((5pi)/(2) + (pi)/(3)) cos (200 pi)`
`=(157cm s^(-1))xxcos (pi)/(3) xx 1= 78.5 cm s^(-1).`
(d) the nodes occur where the amplitude is zero, i.e.,
`sin(1.57 cm^(-1)) x= 0.`
or `((pi)/(2)cm^(-1))x = npi,`
where n is an integer. Thus, x = 2 n cm. the nodes, therefore occur at x = 0,2 cm, 4 cm, 6 cm 8cm and 10cm. Antinodes occure in between them, i.e., at x =1 cm, 3cm, 5 cm, 7 cm and 9cm. The string vibrates is 5 loops.