A wave travelling along a strong is described by y(x,t)=0.005 sin (80.0x-3.0t) in which the numerical constants are in SI units (0.005m, 80.0 rad m^(-1) and 3.0 rad s^( -1)). Calculate (a) the amplitude. (b) the wavelength (c) the period and frequency of the wave. Also , calculate the displacement y of the wave at a distance x=30.0 cm and time t=20 s?
Answers
The wave travelling along the string is described as
y(x,t)=0.005 sin(80x - 3t)
which is of the form
y(x,t)=A sin(kx-ωt)
where A is the amplitude of the wave,
ω the angular frequency and
k the angular wave number
(a) Amplitude of the wave = 0.005m
(b) wavelength is related to wave number by
λ=2π/k
= 2*3.14/ 80=0.0785m=7.85cm
(c) Time period is related to angular frequency by
T=2π/ω= 2*3.14/3= 2.09s
Displacement of the wave at distance 30cm and time 20s can be found by substituting for x and t in the wave equation
y(x,t) = 0.005 sin(80*0.3-3*20)
= 0.005 sin(-36)=0.00495m=4.95mm
We have , y(x,t) = 0.005 sin (80x - 3t) .....(1)
Now we have to find (a) amplitude (b) wavelength (c) time period and displacement of wave at x = 30cm and t = 20sec
Now we know that general equation of a wave is ,
y(x,t) = A sin(kx - wt) .....(2)
where , A is the amplitude of the wave , k is the wave number and w is the angular displacement.
(a) Now , by comparing equation (1) and (2) , we get ,
A = 0.005 m ( amplitude of the wave )
(b) Now , again by comparing (1) and (2) , we get ,
k = 80 / m , where , k = 2π / L ( L is wavelength )
=> L = 2π / k = π / 80 => L = 0.0393 m ( wavelength of the wave )
(c) Now , again equating (1) and (2) ,
w = 3 / sec , where , w = 2π / T ( T is time period )
=> T = 2π / w = 2π / 3
=> T = 2.1 sec ( time period of the wave )
Also , v = 1 / T , where , v is the frequency
=> V = 1 / T = 0.478 Hz ( frequency of the wave )
(*) Now , at x = 3 cm = 0.03 m and t = 20 sec ,
y (0.03 , 20) = 0.005 sin ( 80*0.03 - 3*20 )
=> y = -0.00422 m