A sinusoidal wave is traveling on a string with speed 64.2 cm/s. The displacemen
ID: 1377589 • Letter: A
Question
A sinusoidal wave is traveling on a string with speed 64.2 cm/s. The displacement of the particles of the string at x = 22 cm is found to vary with time according to the equation
y = (8.2 cm) sin[1.2 - (3.5 s-1)t].
The linear density of the string is 8.4 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form
y(x,t) = ym sin(kx - ?t),
what are (c) ym, (d) k, and (e) ?, and (f) the correct choice of sign in front of ?? (g) What is the tension in the string?
Explanation / Answer
Given that
y = (8.2 cm) sin[1.2 - (3.5 s-1)t].
The above equation is in the form of
y(x,t) = ym sin(kx - wt)
a)
Now w =3.5 from the above equation then w=3.5===>2pif =3.5 then frequency f =3.5/2pi =0.557Hz
b)
The wave length of the wave is given by
v =f*lamda
A sinusoidal wave is traveling on a string with speed (v) =64.2 cm/s=0.642m/s
Now the wavelength of the wave is (lamda) =v/f =0.642/0.557 =1.152m
If the equation is in theform of y(x,t) = ym sin(kx - ?t),
Ym is the amplitude of the wave
k =2pi/lamda (Wave number of the wave)
? =w (called angular frequency, measured in radians per second )
The tension in the string is given by v=Sqrt(T/u)
Then T2 =uv2 or Tension in the string (T) =Sqrt(u)*v
u is the linear desnity