A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the system is 2.00 J, find
(a) The force constant of the spring and
(b) The amplitude of the motion.
a) T = 2pi / w so w = 2 pi / T
w = sqrt ( k / m )
so k = w² / m
b) (1/2)* k * xmax ² = 2,00 J
so, xmax = sqrt( 2,00 * 2 / k)
Actually for part a it's;
k=w^2*m
To find the force constant of the spring, we can use the formula:
\( T = 2\pi\sqrt{\frac{m}{k}} \)
where T is the period of the motion, m is the mass of the block, and k is the force constant of the spring.
Given that T = 0.250 s and m = 200 g = 0.2 kg, we can rearrange the formula to solve for k:
\( k = \frac{4\pi^2 m}{T^2} \)
Substituting the given values,
\( k = \frac{4\pi^2 \times 0.2}{(0.250)^2} \)
\( k = \frac{4\pi^2 \times 0.2}{0.0625} \)
\( k = \frac{4 \times 3.14159^2 \times 0.2}{0.0625} \)
\( k \approx 201.06 \, N/m \)
So, the force constant of the spring is approximately 201.06 N/m.
To find the amplitude of the motion, we can use the formula:
\( E = \frac{1}{2} k A^2 \)
where E is the total energy of the system and A is the amplitude of the motion.
Given that E = 2.00 J and k = 201.06 N/m, we can rearrange the formula to solve for A:
\( A = \sqrt{\frac{2E}{k}} \)
Substituting the given values,
\( A = \sqrt{\frac{2 \times 2.00}{201.06}} \)
\( A = \sqrt{\frac{4.00}{201.06}} \)
\( A = \sqrt{0.0199} \)
\( A \approx 0.141 \, m \)
So, the amplitude of the motion is approximately 0.141 m.
To find the force constant of the spring and the amplitude of the motion, we can use the formulas related to the period and the total energy of the simple harmonic motion.
(a) The formula for the period of a mass-spring system is given by:
T = 2π√(m/k)
Where T represents the period, m is the mass of the object, and k is the force constant of the spring.
In this case, we are given that the period is 0.250 seconds and the mass of the block is 200 grams. We need to convert the mass to kilograms:
m = 200 g = 0.2 kg
Plugging in the values into the formula, we have:
0.250 = 2π√(0.2/k)
We can isolate k by squaring both sides of the equation:
(0.250)^2 = (2π)^2 * (0.2/k)
0.0625 = 4π^2 * (0.2/k)
Now, solve for k by rearranging the equation:
k = (4π^2 * 0.2) / 0.0625
Using a calculator to evaluate the expression, we get:
k ≈ 507.96 N/m
So, the force constant of the spring is approximately 507.96 N/m.
(b) The formula for the total energy of a mass-spring system is given by:
E = (1/2) kA^2
Where E represents the total energy and A is the amplitude of the motion.
We are given that the total energy is 2.00 J. Plugging in the values into the formula, we have:
2.00 = (1/2) * 507.96 * A^2
Now, solve for A by rearranging the equation:
A^2 = (2 * 2.00) / 507.96
Evaluate the expression using a calculator:
A^2 ≈ 0.0079
Taking the square root of both sides of the equation, we get:
A ≈ √0.0079
Using a calculator to find the square root, we have:
A ≈ 0.089 meters
So, the amplitude of the motion is approximately 0.089 meters.