The spring 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. Thus began the mass moving up and down with simple harmonic
motion (SHM). In this simple harmonic motion;
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. (3)
c) Determine the speed of the mass when the spring length is 4.30m long. (4)
d) Determine the time it will take the mass to bounce up to the maximum
height and back to its lowest height eight times.
To solve these problems, we need to apply the principles of simple harmonic motion (SHM) and equations for springs.
First, let's solve the questions one by one:
i) At the locations where the velocity of the mass is zero (2):
The velocity of an object undergoing SHM is zero at the extreme points of its motion (i.e., when it reaches the maximum displacement from the equilibrium position). In this case, the equilibrium position is at the unstretched length of the spring, which is 3.00m. The motion is symmetric, so there are two extreme points on either side.
To find the positions where velocity is zero, we need to find the maximum displacement from the equilibrium position. We can calculate this using Hooke's law.
Using Hooke's law: F = -kx
Where F is the restoring force due to the spring, k is the spring constant, and x is the displacement from the equilibrium position.
At the extreme points, the force acting on the mass is only due to gravity (mg), providing the restoring force. So we can write:
mg = kx
Solving for x:
x = mg/k = (7.50kg)(9.8m/s^2)/(75.0N/m) = 0.98m
Therefore, the spring length at the locations where the velocity of the mass is zero is 3.00m ± 0.98m, which gives us two positions: 4.98m and 1.02m.
ii) At the location where its speed is maximized (1):
The speed of the mass is maximized when it passes through the equilibrium position. Therefore, the spring length at this location is the unstretched length of the spring, which is 3.00m.
iii) When its elastic energy is "0J" (1):
The elastic potential energy of a spring is given by the formula: PE = (1/2)kx^2
When the elastic energy is zero, the displacement is also zero. Therefore, the spring length will also be the unstretched length, which is 3.00m.
b) Determine the acceleration of the mass when the spring is 4.20m long (3):
To determine the acceleration, we need to use the equation of motion for SHM:
a = -ω^2x
Where a is the acceleration, ω is the angular frequency (given by ω = √(k/m)), and x is the displacement from the equilibrium position.
Let's calculate it step by step:
First, find ω:
ω = √(k/m) = √(75.0N/m / 7.50kg) = 5.00 rad/s
Now, plug the values into the equation:
a = -(5.00 rad/s)^2 * (4.20m) = -105 m/s^2
Therefore, the acceleration of the mass when the spring is 4.20m long is -105 m/s^2.
c) Determine the speed of the mass when the spring length is 4.30m long (4):
To determine the speed, we can use the equation for velocity in SHM:
v = ω√(A^2 - x^2)
Where v is the velocity, A is the amplitude (maximum displacement), ω is the angular frequency, and x is the displacement from the equilibrium position.
First, let's calculate the maximum displacement (amplitude):
A = 4.50m - 3.00m = 1.50m
Now, plug in the values:
v = (5.00 rad/s)√(1.50m^2 - (4.30m - 3.00m)^2)
= (5.00 rad/s)√(1.50m^2 - 1.30m^2)
= (5.00 rad/s)√(0.25m^2)
= 1.25 m/s
Therefore, the speed of the mass when the spring length is 4.30m long is 1.25 m/s.
d) Determine the time it will take the mass to bounce up to the maximum height and back to its lowest height eight times:
To calculate the time period of the motion, we can use the equation:
T = 2π/ω
First, calculate ω:
ω = √(k/m) = √(75.0N/m / 7.50kg) = 5.00 rad/s
Now, calculate the time period (T):
T = 2π/5.00 rad/s = (2π/5) s
To determine the total time for 8 complete oscillations, multiply the time period by 8:
Total time = (2π/5) s * 8 = (16π/5) s
Therefore, it will take the mass (16π/5) seconds to bounce up to the maximum height and back to its lowest height eight times.