A 450-g block is attached to a spring of spring constant 45 N/m and is in simple harmonic motion on a frictionless surface with an amplitude of 7.5 cm. A second identical block, moving at 12 m/s in line with the spring, strikes the first one when the spring is at its maximum extension. The two blocks stick together. What is the frequency of the resulting oscillation?
85 cm
A 450.0 -g block is attached to a spring of spring constant 45 N/m and is in simple harmonic motion on a frictionless surface with an amplitude of 7.5 cm. A second identical block, moving at 1.2 m/s in line with the spring, strikes the first one when the first block is at its equilibrium position and moving away from the second block. The two blocks stick together. What is the amplitude of the resulting oscillation?
Well, this situation sounds like a real "springer"! Let's see if we can bounce some humor into this physics problem.
First of all, let's calculate the potential energy stored in the spring when it's at its maximum extension. We can use the formula for potential energy in a spring:
Potential Energy = (1/2) * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
In this case, we have an amplitude of 7.5 cm, which is 0.075 meters. So, at maximum extension, the displacement x is 0.075 meters.
The potential energy can be calculated as:
Potential Energy = (1/2) * 45 N/m * (0.075 m)^2
Wait a second... I just realized that we didn't even need to calculate the potential energy! We only need to find the frequency of the resulting oscillation!
Since the two blocks stick together and move as one after collision, we can use the formula for the frequency of a spring-mass system:
Frequency = 1 / (2π) * sqrt(k / m)
where m is the combined mass of the two blocks.
The mass of each block is 450 grams, which is 0.45 kg. Therefore, the combined mass of the two blocks is 0.45 kg * 2 = 0.9 kg.
Now let's calculate the frequency:
Frequency = 1 / (2π) * sqrt(45 N/m / 0.9 kg)
And after a quick calculation, we get:
Frequency ≈ 3.627 Hz
So, the resulting frequency of the oscillation is around 3.627 Hertz. Hopefully, this physics "tumble" wasn't too "bouncing" for you!
To find the frequency of the resulting oscillation, we can use the formula for the angular frequency of simple harmonic motion:
ω = √(k / m)
where ω is the angular frequency, k is the spring constant, and m is the mass.
First, let's convert the amplitude to meters:
amplitude = 7.5 cm = 0.075 m
Since the two blocks stick together after the collision, their combined mass will be:
m = 2 * 450 g = 0.9 kg
Now we can plug in the values into the formula:
ω = √(k / m)
ω = √(45 N/m / 0.9 kg)
ω = √50 rad/s
The frequency (f) of the resulting oscillation can be calculated using the formula:
f = ω / (2π)
f = √50 rad/s / (2π)
f ≈ 3.98 Hz
So the frequency of the resulting oscillation is approximately 3.98 Hz.
To find the frequency of the resulting oscillation, we can use the concepts of conservation of momentum and energy.
First, let's find the initial and final potential energy of the system. The initial potential energy of the system can be calculated using the formula:
PE_initial = (1/2)kx^2
where k is the spring constant and x is the amplitude. Plugging in the values:
PE_initial = (1/2)(45 N/m)(0.075 m)^2 = 0.1519 J
Next, let's find the initial and final kinetic energy of the system. The initial kinetic energy of the system is given by the moving block just before the collision:
KE_initial = (1/2)mv^2
where m is the mass of each block and v is the velocity of the moving block. Plugging in the values:
KE_initial = (1/2)(0.45 kg)(12 m/s)^2 = 32.4 J
Since momentum is conserved in the collision, the final velocity v' of the two blocks after the collision can be calculated using the following equation:
(m1 + m2)v' = m1v1 + m2v2
where m1 and m2 are the masses of the two blocks (which are equal) and v1 and v2 are the initial velocities of the two blocks. Plugging in the values:
(0.45 kg + 0.45 kg)v' = (0.45 kg)(0 m/s) + (0.45 kg)(12 m/s)
(0.9 kg)v' = 5.4 kg⋅m/s
v' = 5.4 kg⋅m/s / 0.9 kg = 6 m/s
Now, let's find the final total energy of the system. The final total energy of the system is given by the combination of kinetic energy and potential energy:
E_final = (1/2)kx'^2 + (1/2)mv'^2
where x' is the maximum extension of the spring after the collision. Since the two blocks stick together, the maximum extension x' is equal to the amplitude of the initial oscillation. Plugging in the values:
E_final = (1/2)(45 N/m)(0.075 m)^2 + (1/2)(0.9 kg)(6 m/s)^2
= 0.1519 J + 16.2 J
= 16.3519 J
Finally, let's find the frequency of the resulting oscillation using the total energy:
E_final = (1/2)kA^2
where A is the amplitude of the oscillation and k is the spring constant. Rearranging the equation to solve for the frequency f:
f = 1 / (2π) * sqrt(k / m) * sqrt(E_final)
Plugging in the values:
f = 1 / (2π) * sqrt(45 N/m / 0.45 kg) * sqrt(16.3519 J)
= 3.88 Hz
Therefore, the frequency of the resulting oscillation is approximately 3.88 Hz.