A diving board oscillates with simple harmonic motion of frequency 4.6 cycles per second. What is the maximum amplitude with which the end of the board can oscillate in order that a pebble placed there does not lose contact with the board during the oscillation?
acceleratoin= -w^2 y
but the max acceleration cannot be greater than g to keep the pebble in place, so
g=-w^2*Amplitude
w=2PIf=9.2PI
solve for amplitude
.03
To determine the maximum amplitude with which the end of the board can oscillate so that a pebble placed there does not lose contact, we need to consider the equilibrium position of the pebble.
In simple harmonic motion, the equilibrium position is the point where the net force acting on the object is zero. In this case, the net force on the pebble must be directed towards the center of the oscillation.
The maximum amplitude occurs when the net force is at its maximum. At this point, the gravitational force acting on the pebble is equal in magnitude but opposite in direction to the centripetal force.
To find the maximum amplitude, we need to equate the gravitational force with the centripetal force:
G = F_c
Where G is the gravitational force and F_c is the centripetal force.
The gravitational force acting on the pebble is given by:
G = m * g
Where m is the mass of the pebble and g is the acceleration due to gravity.
The centripetal force acting on the pebble is given by:
F_c = m * w^2 * A
Where w is the angular frequency (2πf) and A is the amplitude of the oscillation.
Setting G equal to F_c, we have:
m * g = m * w^2 * A
Canceling the mass on both sides of the equation, we get:
g = w^2 * A
Substituting the given frequency f = 4.6 cycles/s and solving for w:
w = 2πf = 2π * 4.6 = 9.2π rad/s
We can now solve for the amplitude A:
A = g / w^2
Substituting g = 9.8 m/s^2 and w = 9.2π rad/s, we get:
A = 9.8 / (9.2π)^2
Using a calculator, we find:
A ≈ 0.0367 m (rounded to 4 decimal places)
Therefore, the maximum amplitude with which the end of the diving board can oscillate so that the pebble does not lose contact is approximately 0.0367 meters.
To determine the maximum amplitude with which the end of the diving board can oscillate, we need to make sure that the pebble maintains contact with the board throughout the motion.
Let's start by understanding simple harmonic motion (SHM). SHM is a type of oscillatory motion where the object moves back and forth around an equilibrium position. In SHM, the displacement (x) of the object from the equilibrium position is a sinusoidal function of time (t), given by the equation:
x(t) = A * cos(ωt + φ)
Where:
- A is the amplitude of the motion (maximum displacement from the equilibrium)
- ω is the angular frequency (related to the frequency of oscillation)
- φ is the phase constant
In this case, the given frequency is 4.6 cycles per second. We can convert this to angular frequency using the formula:
ω = 2πf
Where:
- f is the frequency in cycles per second
- ω is the angular frequency in radians per second
So, ω = 2π * 4.6 = 9.2π rad/s
Now, we need to find the maximum amplitude (A) of the oscillation so that the pebble placed on the end of the board maintains contact. When the pebble loses contact with the board, the acceleration due to gravity (g) should be greater than the centripetal acceleration (ac) required to keep the pebble on the board.
The centripetal acceleration (ac) can be expressed as:
ac = ω² * A
Since the pebble stays in contact throughout the oscillation, we can equate the centripetal acceleration (ac) to the acceleration due to gravity (g):
g = ω² * A
Now, we can solve for the maximum amplitude (A):
A = g / ω²
Substituting the known values, where g is the acceleration due to gravity (approximately 9.8 m/s²) and ω is 9.2π rad/s:
A = 9.8 m/s² / (9.2π rad/s)²
Calculating this value, we get:
A ≈ 9.8 m/s² / (266.96) ≈ 0.0367 meters
Therefore, the maximum amplitude with which the end of the diving board can oscillate, so that the pebble does not lose contact, is approximately 0.0367 meters.