The equation of a transverse wave traveling along a very long
string is given by y = 6.8 sin(0.023px + 3.2pt), where x and y are
expressed in centimeters and t is in seconds. Determine the
following values.
(a) the amplitude
1 cm
string is given by y = 6.8 sin(0.023px + 3.2pt), where x and y are
expressed in centimeters and t is in seconds. Determine the
following values.
(a) the amplitude
1 cm
(b) the wavelength
2 cm
(c) the frequency
3 Hz
(d) the speed
4 cm/s
(e) the direction of propagation of the wave
5
+x -x +y -y
(f) the maximum transverse speed of a particle in the string
6 cm/s
(g) the transverse displacement at x = 3.5 cm when t = 0.26 s
7 cm
Answer
a) Amplitude A = 6.80 cm
b) The wavelenght is
λ=2π/k
=(2π)/(0.023π
rad/cm)
rad/cm)
= 86.95 cm
c) The frequency is
f
=ω/(2π)
=ω/(2π)
=(3.2π)/(2π)
= 1.6 Hz
d) The speed is
v
= ω/k
= ω/k
=
(3.2π rad/s)/(0.023πrad/cm)
(3.2π rad/s)/(0.023πrad/cm)
=
139.13 cm/s
139.13 cm/s
e) -x direction.
f) The maximum speed of a particle in the stringis
vmax=
Aω
Aω
=(6.80
cm)(3.2π rad/s)
cm)(3.2π rad/s)
= 68.36 cm/s
g) The transverse displacement is
y =
6.8 sin(0.023π(3.5cm ) +
4.7π(0.26s))
6.8 sin(0.023π(3.5cm ) +
4.7π(0.26s))
=
- 5.532 cm
- 5.532 cm
