- A mass-spring system undergoes simple harmonic motion. If mass is doubled, what happens to the period?
- A mass-spring system undergoes simple harmonic motion. If the spring constant is doubled, what happens to the period?
- A mass-spring system undergoes simple harmonic motion with amplitude A. What fraction of the system s energy is kinetic energy when the object s position is x=A/2?
- Explain how simple harmonic motion could be used to determine the mass of an astronaut in the apparent weightlessness of an orbiting spacecraft.
- Your car oscillates at a certain frequency after hitting a bump. How does that frequency change as passengers are added?
- What is the net work done by the spring in a simple harmonic oscillator over one complete oscillation cycle? When during the cycle is the spring doing positive work, and when is it doing negative work?
The Correct Answer and Explanation is :
- If the mass is doubled in a mass-spring system undergoing simple harmonic motion (SHM), the period increases.
The period ( T ) of a mass-spring system is given by:
[
T = 2\pi \sqrt{\frac{m}{k}}
]
where ( m ) is the mass and ( k ) is the spring constant.
If the mass is doubled, the period increases by a factor of ( \sqrt{2} ), as the period is directly proportional to the square root of the mass. So, when mass ( m ) is doubled, the period increases by a factor of ( \sqrt{2} ). - If the spring constant is doubled, the period decreases.
The period of the system is inversely proportional to the square root of the spring constant ( k ). Therefore, if the spring constant is doubled, the period ( T ) decreases by a factor of ( \sqrt{2} ), as the spring is stiffer and oscillates faster. - Fraction of the system’s energy that is kinetic energy when the position is ( x = A/2 ):
The total mechanical energy ( E ) in SHM is conserved and is the sum of kinetic energy and potential energy.
The kinetic energy ( K ) and potential energy ( U ) are given by:
[
K = \frac{1}{2} m v^2, \quad U = \frac{1}{2} k x^2
]
At any position ( x ), the total energy is the sum of these two.
At ( x = A/2 ), the potential energy is ( U = \frac{1}{2} k (A/2)^2 = \frac{1}{8} k A^2 ), and the total energy is ( E = \frac{1}{2} k A^2 ).
Therefore, the kinetic energy is the remaining portion, ( K = E – U = \frac{1}{2} k A^2 – \frac{1}{8} k A^2 = \frac{3}{8} k A^2 ).
The fraction of kinetic energy to total energy is:
[
\frac{K}{E} = \frac{\frac{3}{8} k A^2}{\frac{1}{2} k A^2} = \frac{3}{4}
]
Therefore, 75% of the total energy is kinetic energy when ( x = A/2 ). - Using SHM to determine the mass of an astronaut in apparent weightlessness:
In a spacecraft orbiting the Earth, astronauts experience apparent weightlessness, meaning they do not feel their weight due to the absence of contact forces. However, the mass can be determined by using simple harmonic motion.
If a spring is attached to the astronaut and they oscillate in response to a force, the period of oscillation can be used to calculate the mass. The period of a mass-spring system is given by ( T = 2\pi \sqrt{\frac{m}{k}} ), where ( m ) is the mass of the astronaut and ( k ) is the spring constant. By measuring the period ( T ) of the oscillation and knowing the spring constant ( k ), the mass ( m ) of the astronaut can be calculated, even in the apparent weightlessness condition. - Frequency change as passengers are added to a car:
When passengers are added to a car, the total mass of the car increases. The frequency of oscillation is related to the mass of the system, and as the mass increases, the frequency decreases. This is because the frequency ( f ) of a simple harmonic oscillator is given by:
[
f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}
]
where ( k ) is the spring constant and ( m ) is the mass. As the mass increases, the frequency decreases, meaning the car will oscillate at a lower frequency with more passengers. - Net work done by the spring in a simple harmonic oscillator and the direction of work:
In a simple harmonic oscillator, the net work done by the spring over one complete oscillation cycle is zero. This is because the work done by the spring when the object moves in one direction is exactly canceled by the work done in the opposite direction during the return motion.
- Positive work: The spring does positive work when the object is moving towards the equilibrium position (when the object is compressed or stretched and the spring is doing work to bring it back).
- Negative work: The spring does negative work when the object moves away from the equilibrium position, as the spring is resisting the displacement and storing potential energy.
I’ll also generate an image showing the energy exchange in a simple harmonic oscillator over one cycle.
Here’s the image showing the energy exchange in a simple harmonic oscillator over one complete cycle. It illustrates how the kinetic and potential energies change with the object’s position, and that the net work done by the spring over the cycle is zero. Let me know if you need further explanation or adjustments!
