A partridge of mass 5.01kg is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.20s .
What is its speed as it passes through the equilibrium position?
What is its acceleration when it is 0.050 m above the equilibrium position?
When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it?
The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?
This is clearly a problem that can be solved by application of conservation of energy.
Someone will gladly critique your work, when shown.
1.4
To find the speed of the partridge as it passes through the equilibrium position, you can use the equation:
v = 2πA/T
where v is the speed, A is the amplitude (maximum displacement), and T is the period.
In this case, the amplitude is 0.100 m and the period is 4.20 s. Plugging these values into the equation, we get:
v = (2π * 0.100 m) / (4.20 s)
v ≈ 0.1503 m/s
So, the speed of the partridge as it passes through the equilibrium position is approximately 0.1503 m/s.
To find the acceleration of the partridge when it is 0.050 m above the equilibrium position, we can use the equation for the vertical motion of an object attached to a spring:
a = -ω^2x
where a is the acceleration, ω is the angular frequency (ω = 2π / T), and x is the displacement from the equilibrium position.
In this case, x is 0.050 m and the period T is 4.20 s. We can calculate ω as follows:
ω = 2π / T
ω = 2π / 4.20 s
Plugging this value and the displacement into the equation, we get:
a = -(2π / 4.20 s)^2 * 0.050 m
a ≈ -0.0925 m/s^2
So, the acceleration of the partridge when it is 0.050 m above the equilibrium position is approximately -0.0925 m/s^2. The negative sign indicates that the acceleration is directed towards the equilibrium position.
To find the time required for the partridge to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it, we can use the equation for the period of a spring-mass system:
T = 2π * √(m / k)
where T is the period, m is the mass, and k is the spring constant.
In this case, the mass of the partridge is 5.01 kg. However, we need the spring constant to calculate the period accurately. Unfortunately, the spring constant is not provided in the question. Please provide the value of the spring constant to continue with the calculation.
Without the spring constant, we cannot accurately determine the time required for the partridge to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it.
If the motion of the partridge is stopped and then it is removed from the spring, the spring will return to its equilibrium length. Therefore, the spring will shorten by the amount of displacement from its equilibrium position caused by the partridge's weight.
The displacement caused by the partridge's weight can be calculated using Hooke's law:
F = kx
where F is the force applied by the partridge's weight, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, the force F is given by:
F = mg
where m is the mass (5.01 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we get:
F = (5.01 kg) * (9.8 m/s^2)
Using Hooke's law, we can find the displacement x:
kx = (5.01 kg) * (9.8 m/s^2)
Since the spring is ideal (negligible mass), the displacement will be equal to the extension of the spring:
x = (5.01 kg) * (9.8 m/s^2) / k
Therefore, to determine how much the spring shortens, we need the value of the spring constant (k). Please provide the value of the spring constant to continue with the calculation.
To find the speed of the partridge as it passes through the equilibrium position, we can use the equation for simple harmonic motion:
v = ωA
where v is the speed, ω is the angular frequency, and A is the amplitude of the motion.
First, we need to find the angular frequency. The period T of the motion is given as 4.20 s. The angular frequency ω is related to the period by the equation:
ω = (2π) / T
Plugging in the values, we have:
ω = (2π) / 4.20 s ≈ 1.5 rad/s
Next, we need to find the amplitude A, which is the maximum displacement from the equilibrium position. The question tells us that the partridge is pulled down 0.100 m below its equilibrium position. Therefore, the amplitude is 0.100 m.
Now, we can calculate the speed:
v = ωA = 1.5 rad/s * 0.100 m ≈ 0.15 m/s
Therefore, the speed of the partridge as it passes through the equilibrium position is approximately 0.15 m/s.
To find the acceleration when the partridge is 0.050 m above the equilibrium position, we can use the equation for acceleration in simple harmonic motion:
a = -ω²x
where a is the acceleration, ω is the angular frequency, and x is the displacement from the equilibrium position.
First, we find the angular frequency ω as we did in the previous calculation:
ω = 1.5 rad/s
Next, we plug in the values:
a = - (1.5 rad/s)² * 0.050 m ≈ -0.11 m/s²
Therefore, the acceleration when the partridge is 0.050 m above the equilibrium position is approximately -0.11 m/s². The negative sign indicates that the acceleration is directed downward.
To find the time required for the partridge to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it, we can use the equation for the period T of the motion:
T = 2π / ω
where T is the period, ω is the angular frequency.
We already have the angular frequency ω:
ω = 1.5 rad/s
Plugging in the values:
T = 2π / 1.5 rad/s ≈ 4.19 s
Therefore, the time required for the partridge to move from a point 0.050 m below its equilibrium position to a point 0.050 m above it is approximately 4.19 seconds.
Finally, to find how much the spring shortens when the motion of the partridge is stopped and it is removed, we need to use Hooke's Law for springs.
Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position:
F = -kx
where F is the force, k is the spring constant, and x is the displacement.
The displacement x is given as 0.100 m (it is the same as the amplitude from the first question). However, in this case, the force is equal to the weight of the partridge, which is given by the equation:
F = mg
where m is the mass of the partridge and g is the acceleration due to gravity.
The question gives the mass of the partridge as 5.01 kg.
Therefore, we have:
mg = kx
Solving for k, we have:
k = mg / x
Plugging in the values:
k = (5.01 kg * 9.8 m/s²) / 0.100 m ≈ 49 N/m
Now, we need to find the change in length of the spring. The spring is in equilibrium when the partridge is not moving, so the displacement x in Hooke's Law is zero. We can find the change in length ΔL of the spring when it is removed:
ΔL = x - (mg / k)
Plugging in the values:
ΔL = 0.100 m - (5.01 kg * 9.8 m/s²) / (49 N/m) ≈ -0.100 m
The negative sign indicates that the spring shortens by approximately 0.100 m when the partridge is removed.
Therefore, the spring shortens by approximately 0.100 m.