Question
A weight of mass m is at rest at O when suspended from a spring, as shown
in figure 1. 0. The energy applied (E) of pulling down the spring is
combination of potential energy (PE) and kinetic energy (KE). When released,
the spring oscillates between positions A and B.
2
(i)
Given that the amplitude A is equal to the maximum displacement, Xmax
Given that the amplitude A is equal to the maximum displacement, Xmax
i.e A =. Xmax, show that the velocity of the object is given by the equation.
�
� =√𝑘
𝑚
(𝐴2 −𝑥2)
in figure 1. 0. The energy applied (E) of pulling down the spring is
combination of potential energy (PE) and kinetic energy (KE). When released,
the spring oscillates between positions A and B.
2
(i)
Given that the amplitude A is equal to the maximum displacement, Xmax
Given that the amplitude A is equal to the maximum displacement, Xmax
i.e A =. Xmax, show that the velocity of the object is given by the equation.
�
� =√𝑘
𝑚
(𝐴2 −𝑥2)
Answers
GPT-4o mini
To show that the velocity of an oscillating mass attached to a spring is given by the equation
\[
v = \sqrt{\frac{k}{m} (A^2 - x^2)}
\]
we start by considering the potential energy (PE) and kinetic energy (KE) in the system.
1. **Potential Energy in the Spring:**
The potential energy stored in a stretched or compressed spring is given by the equation:
\[
PE = \frac{1}{2} k x^2
\]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.
2. **Kinetic Energy of the Mass:**
The kinetic energy (KE) of the mass \( m \) with a velocity \( v \) is given by:
\[
KE = \frac{1}{2} m v^2
\]
3. **Total Mechanical Energy:**
In a spring-mass system without any damping (no energy loss), the total mechanical energy (E) is conserved and is the sum of the potential and kinetic energy:
\[
E = PE + KE
\]
At the maximum displacement \( x = A \) (where the spring is fully stretched or compressed), all the energy is potential energy, so:
\[
E = \frac{1}{2} k A^2
\]
At any other position \( x \), the total energy can be expressed as:
\[
E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2
\]
4. **Setting the Total Energy Equations Equal:**
Since energy is conserved, we can equate the two expressions for total energy:
\[
\frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2} m v^2
\]
We can simplify by multiplying through by 2 to get rid of the fractions:
\[
k A^2 = k x^2 + m v^2
\]
5. **Solving for the Velocity:**
Rearranging this equation, we have:
\[
m v^2 = k A^2 - k x^2
\]
Factor out \( k \):
\[
m v^2 = k (A^2 - x^2)
\]
Now, divide both sides by \( m \):
\[
v^2 = \frac{k}{m} (A^2 - x^2)
\]
Taking the square root of both sides gives us the velocity:
\[
v = \sqrt{\frac{k}{m} (A^2 - x^2)}
\]
This shows that the velocity \( v \) of the mass as it oscillates is given by the equation
\[
v = \sqrt{\frac{k}{m} (A^2 - x^2)}
\]
as required.
\[
v = \sqrt{\frac{k}{m} (A^2 - x^2)}
\]
we start by considering the potential energy (PE) and kinetic energy (KE) in the system.
1. **Potential Energy in the Spring:**
The potential energy stored in a stretched or compressed spring is given by the equation:
\[
PE = \frac{1}{2} k x^2
\]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.
2. **Kinetic Energy of the Mass:**
The kinetic energy (KE) of the mass \( m \) with a velocity \( v \) is given by:
\[
KE = \frac{1}{2} m v^2
\]
3. **Total Mechanical Energy:**
In a spring-mass system without any damping (no energy loss), the total mechanical energy (E) is conserved and is the sum of the potential and kinetic energy:
\[
E = PE + KE
\]
At the maximum displacement \( x = A \) (where the spring is fully stretched or compressed), all the energy is potential energy, so:
\[
E = \frac{1}{2} k A^2
\]
At any other position \( x \), the total energy can be expressed as:
\[
E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2
\]
4. **Setting the Total Energy Equations Equal:**
Since energy is conserved, we can equate the two expressions for total energy:
\[
\frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2} m v^2
\]
We can simplify by multiplying through by 2 to get rid of the fractions:
\[
k A^2 = k x^2 + m v^2
\]
5. **Solving for the Velocity:**
Rearranging this equation, we have:
\[
m v^2 = k A^2 - k x^2
\]
Factor out \( k \):
\[
m v^2 = k (A^2 - x^2)
\]
Now, divide both sides by \( m \):
\[
v^2 = \frac{k}{m} (A^2 - x^2)
\]
Taking the square root of both sides gives us the velocity:
\[
v = \sqrt{\frac{k}{m} (A^2 - x^2)}
\]
This shows that the velocity \( v \) of the mass as it oscillates is given by the equation
\[
v = \sqrt{\frac{k}{m} (A^2 - x^2)}
\]
as required.