A long straight horizontal bar vibrates in a ver
tical direction with a frequency of 1.50 Hz on the
surface of a large
tank of water. The vibrations of the bar cause plan
e waves to travel across the surface of the water.
The waves are
described by the equation
−
λ
π
=
)
(
2
sin
vt
x
A
y
where
t
is the time and
x
is the distance from the bar. The velocity
of the waves is 12.00 cm/s and the wave amplitude is 2.00 cm.
(i) An insect (the sort with such a tiny mass that
they can actually walk on the surface of a lake) is
30.0 cm from
the bar. Calculate the vertical displacement of the
insect at the time t = 1.00 s.
To find the vertical displacement of the insect at a given time (t = 1.00 s), we need to substitute the given values into the equation and solve for y.
Given:
Frequency of vibration (f) = 1.50 Hz
Wave velocity (v) = 12.00 cm/s
Wave amplitude (A) = 2.00 cm
Distance from the bar (x) = 30.0 cm
Time (t) = 1.00 s
The equation for the wave is:
y = A*sin(2πft - λx)
We can calculate the wavelength (λ) using the formula:
λ = v/f
Substituting the given values:
λ = 12.00 cm/s / 1.50 Hz
λ = 8.00 cm
Now we have all the necessary values to calculate the vertical displacement at t = 1.00 s.
y = A*sin(2πft - λx)
Substituting the values:
y = 2.00 cm * sin(2π*1.50 Hz*1.00 s - 8.00 cm*30.0 cm)
Now we can evaluate this expression:
y = 2.00 cm * sin(2π*1.50 - 8.00 cm*30.0 cm)
y = 2.00 cm * sin(3π - 240.00 cm)
y = 2.00 cm * sin(3π - 240.00 cm)
Using a calculator, we find:
y ≈ -2.00 cm
Therefore, at t = 1.00 s, the vertical displacement of the insect 30.0 cm from the bar is approximately -2.00 cm.