The equation of transverse wave on a rope is
y(x, t) = 5sin (4.0t − 0.02x)
where y and x are measured in cm and t is expressed in second. Calculate the
maximum speed of a particle on the rope.
Answers
20cm/s should be the answer
Answer:
The wave function for the transverse wave on the rope can be written as follows:
y(x,t) = A sin(\omega t - kx) = 5 sin(4.0 t - 0.02x) ,y(x,t)=Asin(ωt−kx)=5sin(4.0t−0.02x),
here, A = 0.05 mA=0.05m is the amplitude of the transverse wave, \omega = 4.0 \dfrac{rad}{s}ω=4.0
s
rad
is the angular frequency of the transverse wave and k = 0.02 \dfrac{rad}{cm} \cdot \dfrac{100 cm}{1 m} = 2 \dfrac{rad}{m}k=0.02
cm
rad
⋅
1m
100cm
=2
m
rad
is the wavenumber.
By the definition, the velocity is the derivative of position with respect to time:
v = \dfrac{dy}{dt}.v=
dt
dy
.
So, we can write the speed of a particle on the rope as follows:
v(x, t) = \dfrac{dy(x,t)}{dt} = \dfrac{d}{dt}(A sin(\omega t - kx)) = A \omega cos(\omega t - kx).v(x,t)=
dt
dy(x,t)
=
dt
d
(Asin(ωt−kx))=Aωcos(ωt−kx).
As we can see from the formula, the speed of a particle on the rope is maximum when cos(\omega t - kx) = \pm 1cos(ωt−kx)=±1.
Therefore,
v_{max} = A \omega = 0.05 m \cdot 4.0 \dfrac{rad}{s} = 0.2 \dfrac{m}{s}.v
max
=Aω=0.05m⋅4.0
s
rad
=0.2
s
m
Explanation: