The spring shown in the image below has an unstretched length of 3.00m, and a spring constant of 75.0N/m. The mass of m=7.50kg is attached to the end of the spring, and the mass is pulled down to stretch the spring to 4.50m in length. This began the mass moving up and down with simple harmonic motion (SHM). In this simple harmonic motion;
a) What is the spring length;
i) At the locations where the velocity of the mass is zero (2)
ii) At the location where its speed is maximized (1)
iii) when its elastic energy is “0J” (1)
b) Determine the acceleration of the mass when the spring is 4.20m long.
c) Determine the speed of the mass when the spring length is 4.30m long.
d) Determine the time it will take the mass to bounce up to the maximum
height and back to its lowest height eight times.
F = k x
omega = w = 2 pi f = 2 pi/T = sqrt (k/m) = sqrt 10
say g = 10 m/s^2
then for x of center of sinusoidal motion = m g/k = 7.5 * 10 / 75 = 1 meter of stretch due purely to gravity
so
answer to first part of first question is 4 meters plus or minus 0.5meters
answer to second part of first question is 3 + 1 = 4 meters spring length of center of sinusoidal motion where speed is max
then
say x = 4.0 + 0.5 sin w t = 4.0 + 0.5 sin [ (sqrt 10 )* t]
then v = 0.5 sqrt 10 * cos [ (sqrt 10 )* t]
and a = - 5 sin [ (sqrt 10 )* t]
f = omega/2 pi = sqrt 10/ 2pi
so
T = 1/f = 2 pi / sqrt 10
To answer these questions, we need to understand the concepts of simple harmonic motion (SHM) and the behavior of a mass-spring system. We'll go step by step to solve each part of the question:
a) What is the spring length;
i) At the locations where the velocity of the mass is zero (2)
ii) At the location where its speed is maximized (1)
iii) When its elastic energy is "0J" (1)
To answer these questions, we need to find the equilibrium position of the mass-spring system. In simple harmonic motion, the equilibrium position is where the net force acting on the mass is zero.
1) The equilibrium position is the actual length of the spring when there is no external force acting on it. In this case, the spring length is given as 3.00 m.
2) At the positions where the velocity of the mass is zero, the spring is momentarily at rest and the elongations are equal. Since the spring is stretched to 4.50 m, and the unstretched length is 3.00 m, the elongations on both sides are (4.50 - 3.00) / 2 = 0.75 m. Thus, the spring length at these positions is 3.00 m + 0.75 m = 3.75 m.
3) When the elastic energy is zero, the spring is at its maximum compression or elongation. In this case, the spring is stretched to 4.50 m, so the maximum elongation is (4.50 - 3.00) = 1.50 m. Therefore, the spring length is 3.00 m + 1.50 m = 4.50 m.
b) To determine the acceleration of the mass when the spring is 4.20 m long, we'll use the equation for the acceleration in SHM: a = -(ω^2) * x. Here, ω is the angular frequency, and x is the displacement from equilibrium position.
The angular frequency (ω) can be calculated using the formula: ω = √(k / m), where k is the spring constant and m is the mass.
Given k = 75.0 N/m and m = 7.50 kg, we have ω = √(75.0 N/m / 7.50 kg) = √10.0 rad/s.
Now, the displacement from equilibrium position is x = 4.20 m - 3.00 m = 1.20 m.
So, the acceleration can be calculated as follows:
a = -(√10.0 rad/s)^2 * 1.20 m = -10.0 m/s^2 (negative sign indicates opposite direction of displacement).
Therefore, the acceleration of the mass when the spring is 4.20 m long is -10.0 m/s^2.
c) To determine the speed of the mass when the spring length is 4.30 m long, we need to calculate the maximum speed using the equation: vmax = ω * A, where vmax is the maximum speed, ω is the angular frequency, and A is the amplitude (maximum displacement).
We already calculated ω as √10.0 rad/s. The amplitude can be calculated as A = (4.50 m - 3.00 m) / 2 = 0.75 m.
Therefore, vmax = √10.0 rad/s * 0.75 m = √7.50 m^2/s^2 = 2.74 m/s (rounded to two decimal places).
Therefore, the speed of the mass when the spring length is 4.30 m long is approximately 2.74 m/s.
d) To determine the time it will take the mass to bounce up to the maximum height and back to its lowest height eight times, we need the time period (T) of the system.
The time period (T) can be calculated using the formula: T = 2π / ω, where ω is the angular frequency.
Substituting ω = √10.0 rad/s, we have T = 2π / √10.0 rad/s = 2π / 3.16 rad/s = 1.99 s (rounded to two decimal places).
The number of oscillations is given as eight. As each oscillation consists of one complete cycle from maximum height to lowest height, there will be eight complete cycles.
Therefore, the total time taken will be 8 * T = 8 * 1.99 s = 15.92 s (rounded to two decimal places).
Therefore, it will take the mass approximately 15.92 seconds to bounce up to the maximum height and back to its lowest height eight times.