A 0.5 kg air-track car is attached to the end of a horizontal spring of constant k = 20 N/m. The car is displaced 15 cm from its equilibrium point and released.
a) What is the car's maximum speed?
b) What is the car's maximum acceleration?
c) What is the frequency f of the car's oscillation?
I need help please! Do it step by step and then substitute the numbers into that equation
determine max speed from
maxKE=maxSpringPE
for max acceleration
1/2 m v^2=1/2 kx^2
2v v'=2kx v
v' acceleration=kx/m
period=2PIsqrt(mass/k)
and freq= 1/period
Well, well, well! Let's take this step by step and have some fun with it!
a) First, let's find the car's maximum speed, v_max. We can use the fact that the total mechanical energy is conserved, which means that the sum of the kinetic energy and potential energy remains constant throughout the motion.
At the maximum displacement, the car's potential energy is at its maximum and kinetic energy is zero. At the equilibrium position, the potential energy is zero and the kinetic energy is at its maximum.
So, let's equate these energies:
Potential Energy at maximum displacement = Kinetic Energy at equilibrium
Now, the potential energy of a spring is given by the equation U = (1/2)kx^2, where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.
For the potential energy at maximum displacement, we have:
U = (1/2)kx^2
= (1/2)(20 N/m)(0.15 m)^2
= 0.225 J (joules)
Since the kinetic energy at equilibrium is equal to the maximum potential energy, we have:
(1/2)mv_max^2 = 0.225 J
Solving for v_max:
v_max^2 = (2 * 0.225 J) / 0.5 kg
v_max = √(2 * 0.225 J / 0.5 kg)
v_max = √(0.9 J / 0.5 kg)
v_max = √1.8 J/kg
v_max ≈ 1.34 m/s
So, the car's maximum speed is approximately 1.34 m/s.
b) Now, let's find the car's maximum acceleration, a_max. The maximum acceleration occurs when the car is passing through its equilibrium position, where the net force is at its maximum.
The net force acting on the car is given by the equation F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.
At the maximum displacement, the force is maximum in magnitude and directed towards the equilibrium position. So, F_max = kx.
Using Newton's second law, F = ma, we can solve for the maximum acceleration, a_max:
F_max = kx
ma_max = kx
a_max = (k / m) * x
a_max = (20 N/m) / 0.5 kg * (0.15 m)
a_max = 20 N/m / 0.5 kg * 0.15 m
a_max = 6 m/s^2
So, the car's maximum acceleration is 6 m/s^2.
c) Lastly, let's find the frequency, f, of the car's oscillation. The frequency is given by the equation f = (1 / 2π) * √(k / m), where f is the frequency, k is the spring constant, and m is the mass.
Substituting the values:
f = (1 / 2π) * √(20 N/m / 0.5 kg)
f = (1 / 2π) * √(40 N/kg)
f ≈ (1 / 2π) * √(40 Hz)
f ≈ (1 / 2π) * 6.32 Hz
f ≈ 1 Hz
So, the frequency of the car's oscillation is approximately 1 Hz.
And there you have it, my friend! The car's maximum speed is about 1.34 m/s, its maximum acceleration is 6 m/s^2, and its frequency is approximately 1 Hz. Keep those questions coming, and I'll keep the laughter rolling!
Sure! Let's go step by step to solve this problem.
Step 1: Find the spring constant k.
The problem states that the spring constant is k = 20 N/m.
Step 2: Find the maximum speed.
The maximum speed of the car can be found using the equation: vmax = Aω, where A is the amplitude of the oscillation and ω is the angular frequency. In this case, the amplitude is the displacement from the equilibrium position, which is 15 cm or 0.15 m.
Therefore, A = 0.15 m.
Step 3: Find the angular frequency ω.
The angular frequency ω can be calculated using the formula: ω = sqrt(k/m), where m is the mass of the car. In this case, the mass of the car is 0.5 kg.
Therefore, m = 0.5 kg.
Plugging in the values, ω = sqrt(20 N/m / 0.5 kg) = sqrt(40) = 2 sqrt(10) rad/s.
Step 4: Calculate the maximum speed.
Substituting the values of A and ω into the equation vmax = Aω, we get:
vmax = (0.15 m) x (2 sqrt(10) rad/s) = 0.3 sqrt(10) m/s.
Therefore, the car's maximum speed is 0.3 sqrt(10) m/s.
Step 5: Find the maximum acceleration.
The maximum acceleration of the car can be found using the equation: amax = Aω^2. In this case:
amax = (0.15 m) x (2 sqrt(10) rad/s)^2 = 0.6 x 10 m/s^2 = 6 m/s^2.
Therefore, the car's maximum acceleration is 6 m/s^2.
Step 6: Calculate the frequency.
The frequency f of the car's oscillation can be calculated using the equation: f = ω / (2π). In this case:
f = (2 sqrt(10) rad/s) / (2π) = sqrt(10) / π Hz.
Therefore, the car's frequency is sqrt(10) / π Hz.
To calculate the numerical values, substitute sqrt(10) ≈ 3.16 and π approximately as 3.14 into the equations.
a) The car's maximum speed ≈ 0.3 x 3.16 ≈ 0.948 m/s.
b) The car's maximum acceleration ≈ 6 m/s^2.
c) The car's frequency f ≈ 3.16 / 3.14 ≈ 1.006 Hz.
Sure, let's break down the problem step by step and solve it using the given information.
a) The maximum speed of the car can be found using the equation for simple harmonic motion:
v_max = Aω
Where:
v_max = maximum speed
A = amplitude (displacement from the equilibrium point)
ω = angular frequency
In this case, the amplitude is given as 15 cm, which we need to convert to meters by dividing by 100:
A = 15 cm / 100 = 0.15 m
To find the angular frequency, we can use the formula:
ω = √(k/m)
Where:
k = spring constant = 20 N/m
m = mass of the car = 0.5 kg
Plugging in the values, we get:
ω = √(20 N/m / 0.5 kg) = √(40 m/s^2 / kg) = √(40) ≈ 6.32 rad/s
Now we can substitute these values into the equation to find the maximum speed:
v_max = Aω = 0.15 m * 6.32 rad/s = 0.948 m/s
So, the car's maximum speed is approximately 0.948 m/s.
b) The maximum acceleration can be found using the equation:
a_max = Aω^2
Using the same values for A and ω as before, we can calculate:
a_max = 0.15 m * (6.32 rad/s)^2 = 0.15 m * 39.9 m^2/s^2 = 5.985 m/s^2
Therefore, the car's maximum acceleration is approximately 5.985 m/s^2.
c) The frequency of the car's oscillation, f, can be found using the formula:
f = ω / (2π)
Plugging in the value of ω, we get:
f = 6.32 rad/s / (2π) = 6.32 rad/s / 6.28 ≈ 1.005 Hz
So, the frequency of the car's oscillation is approximately 1.005 Hz.
I hope this explanation helps! Let me know if there's anything else I can assist you with.