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) Let A be the amplitude of vibration of the particle
`A = 3 cm sin (1.57 cm^(-1)) x`
At x = 3.66 cm
`A = 3 sin[(pi)/(2) xx 3.66] (1.57 = (pi)/(2))`
`= 3 sin [(3pi)/(2) + (pi)/(3)]`
`= "3 cos" (pi)/(3) = (3)/(2) = 1.5` cm
From (i), we can say
`k = 1.57 cm6(-1), omega = 314 s^(-1)`
As `k = (2pi)/(lambda)`
`therefore" "lambda = (2pi)/(k) = (2 xx 3.14)/(1.57)= 4` cm
As `omega = 2 pi v1 because` frequency `= v = (omega)/(2pi) = (314)/(2 xx 3.14) = 50 s^(-1)`
Speed of wave `= v lambda = (50) xx 4 = 200` cm/s = 2 m/s
(c) `upsilon` = velocity of particle at x = 3.66 cm at t = 4s.
`upsilon = (dy)/(dt) = (3 cm) sin (1.57 cm^(-1)) x (314) cos (314 s^(-1))t`
`= (942 cm) sin (1.57 cm^(-1)) x cos (314 s^(-1))t" "...(ii)`
Put x = 3.66 cm and t = 4 s in (ii)
`upsilon = (942 cm) sin ((3pi)/(2) + (pi)/(3)) cos (200 pi)`
= 47.1 cm/s
(d) The nodes will be formed when amplitude will be zero, i.e. `sin (1.57 cm^(-1)) x = 0`
`((pi)/(2) cm^(-1))x = n pi` (n is an integer)
`therefore` we get x = 2n
The nodes will be, therefore, formed at x = 0, 2, 4, 6, 8 cm and antinodes will be formed at x = 1 cm 3 cm, 5 cm, 7 cm.
As 4 antinodes are formed. Therefore, string will vibrate in 4 loops.