A small piece of cork in a ripple tank oscillates up and down as ripples pass it. If the ripples travel at 0.20m/s, have a wavelength of 15mm and an amplitude of 5.0mm, what is the maximum velocity of the cork?
y means lambda
v= fy
y- 15mm= 0.015m
A- 5mm= 0.005m
0.2= f*0.015
f= 0.2/0.015= 40/3 Hz
w=2πf
w= 2*22/7*40/3= 83.81
V(max)= wA
= 83.81*0.005= 0.42m/s
The answer is 0.419m/s
.44 m/s
To find the maximum velocity of the cork, we need to consider the relationship between the wave properties and the motion of the cork.
Given information:
- Wave speed (v) = 0.20 m/s
- Wavelength (λ) = 15 mm = 0.015 m (since 1 mm = 0.001 m)
- Amplitude (A) = 5.0 mm = 0.005 m (since 1 mm = 0.001 m)
The maximum velocity of the cork occurs when it reaches the maximum displacement from its equilibrium position, which is equal to the amplitude of the wave.
The formula relating wave speed, frequency, and wavelength is:
v = f * λ
We can rearrange the equation to solve for frequency (f):
f = v / λ
Since frequency (f) is the number of oscillations per second, it is also equal to the number of complete wave cycles per second.
To find the maximum velocity (V_max) of the cork, we can use the formula:
V_max = 2πfA
Now, let's substitute the given values into the equations to find the maximum velocity of the cork:
1. Calculate frequency (f):
f = v / λ
f = 0.20 m/s / 0.015 m = 13.33 Hz
2. Calculate maximum velocity (V_max):
V_max = 2πfA
V_max = 2π * 13.33 Hz * 0.005 m
V_max ≈ 0.42 m/s
Therefore, the maximum velocity of the cork is approximately 0.42 m/s.