Energy in Simple Harmonic Motion: Videos & Practice Problems

In a mass-spring system undergoing simple harmonic motion, energy oscillates between kinetic energy and elastic potential energy. At maximum amplitude, elastic potential energy is maximized while kinetic energy is zero. Conversely, at equilibrium, kinetic energy is maximized and elastic potential energy is zero. The conservation of mechanical energy can be expressed as 1/2k(a)2 = 1/2k(x)2 + 1/2mv^2. This relationship allows for the calculation of velocity at any position.

Energy in Simple Harmonic Motion

In a mass-spring system undergoing simple harmonic motion, energy is continuously exchanged between kinetic energy and elastic potential energy. The kinetic energy, associated with the motion of the mass, is given by the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity. The elastic potential energy, related to the deformation of the spring, is expressed as \( PE = \frac{1}{2} kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

In the absence of friction, the total mechanical energy of the system remains constant, allowing us to apply the principle of conservation of energy. This principle can be illustrated by examining the energy at three key positions: the amplitude (maximum displacement) and the equilibrium position (where \( x = 0 \)). At the amplitude, the elastic potential energy is maximized, calculated as \( \frac{1}{2} kA^2 \), while the kinetic energy is zero since the mass momentarily stops. Conversely, at the equilibrium position, the elastic potential energy is zero, and the kinetic energy reaches its maximum value, represented as \( \frac{1}{2} mv_{\text{max}}^2 \).

For any other position \( x \), the total mechanical energy can be expressed as the sum of kinetic and potential energies: \( \frac{1}{2} kA^2 = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 \). This equation allows us to relate the energies at different points in the motion. By manipulating this equation, we can derive the velocity as a function of position, given by:

\( v(x) = \sqrt{\frac{k}{m}(A^2 - x^2)} \)

In practical applications, such as finding the maximum speed or the speed at a specific position, we can utilize the conservation of energy equation. For example, if we have a mass of 5 kg, a spring constant \( k = 30 \, \text{N/m} \), and an amplitude of \( 0.4 \, \text{m} \), we can calculate the maximum speed \( v_{\text{max}} \) using the relationship:

\( v_{\text{max}} = A \sqrt{\frac{k}{m}} \)

Substituting the values, we find \( v_{\text{max}} \approx 0.98 \, \text{m/s} \). To find the speed at a specific position, such as \( x = -0.2 \, \text{m} \), we can use the derived velocity function:

\( v(-0.2) = \sqrt{\frac{30}{5}(0.4^2 - (-0.2)^2)} \approx 0.85 \, \text{m/s} \)

Finally, the total mechanical energy of the system can be calculated using any of the energy expressions, confirming that it remains constant throughout the motion. For instance, using the potential energy at maximum displacement yields a total mechanical energy of \( 2.40 \, \text{J} \), consistent with calculations using kinetic energy at maximum speed.

Example

In this scenario, we are analyzing a mass-spring system where a 0.25 kg mass oscillates with a period of 3.2 seconds. At a specific position of 0.4 meters, the speed of the mass is 5 m/s. Our goal is to determine the amplitude of the oscillation, denoted as \( A \).

To find the amplitude, we can utilize the relationship between velocity, position, and amplitude in harmonic motion. The relevant equation is:

\( v_x = \omega \sqrt{A^2 - x^2} \)

Here, \( v_x \) is the velocity at position \( x \), and \( \omega \) is the angular frequency. First, we need to calculate \( \omega \) using the period:

\( \omega = \frac{2\pi}{T} \)

Substituting the given period:

\( \omega = \frac{2\pi}{3.2} \approx 1.96 \, \text{rad/s} \)

Now, substituting the known values into the velocity equation:

\( 5 = 1.96 \sqrt{A^2 - 0.4^2} \)

Rearranging gives:

\( \sqrt{A^2 - 0.16} = \frac{5}{1.96} \approx 2.55 \)

Squaring both sides results in:

\( A^2 - 0.16 = 2.55^2 \)

Calculating \( 2.55^2 \) yields approximately 6.50, leading to:

\( A^2 = 6.50 + 0.16 = 6.66 \)

Taking the square root gives:

\( A \approx 2.58 \, \text{m} \)

Next, we need to calculate the total mechanical energy of the system. The mechanical energy \( E \) in a mass-spring system can be expressed as:

\( E = \frac{1}{2} k A^2 \)

To find \( k \), we can use the relationship between angular frequency and spring constant:

\( \omega = \sqrt{\frac{k}{m}} \)

Rearranging gives:

\( k = m \omega^2 \)

Substituting the known values:

\( k = 0.25 \times (1.96)^2 \approx 0.96 \, \text{N/m} \)

Now substituting \( k \) and \( A \) into the energy equation:

\( E = \frac{1}{2} \times 0.96 \times (2.58)^2 \approx 3.20 \, \text{J} \)

In summary, the amplitude of the oscillation is approximately 2.58 meters, and the total mechanical energy of the system is about 3.20 joules.

A block of mass 0.300 kg is attached to a spring. At x = 0.240 m, its acceleration is a_x= -12.0 m/s² and its velocity is v_x=4.00 m/s. What are the system's (a) force constant k and (b) amplitude of motion?

• A. 9 N/m; .35 m
• B. 15 N/m; .61 m
• C. 15 N/m; 1.22 m
• D. 9 N/m; .11 m

Answer: (B)

Example

In the study of oscillations, understanding the relationship between amplitude and various parameters is crucial. When the amplitude of an oscillation increases, it does not affect the period of oscillation. The period, denoted as \( T \), is related to the angular frequency \( \omega \) by the equation:

\( T = \frac{2\pi}{\omega} \)

Here, \( \omega \) is determined by the properties of the system, such as the spring constant \( k \) and mass \( m \), and is independent of amplitude. Therefore, an increase in amplitude does not lead to an increase in the period of oscillation.

Conversely, the maximum acceleration \( a_{\text{max}} \) is directly influenced by amplitude. The formula for maximum acceleration is given by:

\( a_{\text{max}} = \frac{k}{m} \cdot A \)

where \( A \) is the amplitude. As amplitude increases, \( a_{\text{max}} \) also increases, confirming that this statement is correct.

Next, the maximum speed \( v_{\text{max}} \) is also affected by amplitude, expressed as:

\( v_{\text{max}} = A \cdot \omega \)

Since \( \omega \) remains constant with changes in amplitude, an increase in amplitude directly results in an increase in \( v_{\text{max}} \).

When considering maximum kinetic energy \( K_{\text{max}} \), it is defined as:

\( K_{\text{max}} = \frac{1}{2} v_{\text{max}}^2 \)

Since \( v_{\text{max}} \) increases with amplitude, \( K_{\text{max}} \) also increases as a result.

The maximum potential energy \( U_{\text{max}} \) in a spring system is given by:

\( U_{\text{max}} = \frac{1}{2} k A^2 \)

As amplitude increases, \( U_{\text{max}} \) increases as well, indicating a direct relationship.

Finally, the total mechanical energy \( E_{\text{max}} \) of the system, which is the sum of kinetic and potential energy, is expressed as:

\( E_{\text{max}} = \frac{1}{2} k A^2 \)

Thus, an increase in amplitude leads to an increase in total mechanical energy, confirming that all aspects of energy in the system are positively correlated with amplitude.

Here’s what students ask on this topic:

What is the relationship between kinetic energy and potential energy in simple harmonic motion?

In simple harmonic motion, the total mechanical energy is conserved and oscillates between kinetic energy (KE) and elastic potential energy (PE). At maximum amplitude, the potential energy is maximized, and kinetic energy is zero. Conversely, at the equilibrium position, kinetic energy is maximized, and potential energy is zero. The conservation of mechanical energy can be expressed as:

$\frac{1}{2}k{a}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

This equation shows that the sum of kinetic and potential energy remains constant throughout the motion.

How do you calculate the maximum speed in a mass-spring system undergoing simple harmonic motion?

To calculate the maximum speed (v_max) in a mass-spring system undergoing simple harmonic motion, you can use the conservation of energy principle. The maximum speed occurs at the equilibrium position where all the energy is kinetic. The equation is:

$v$_max = a√km

Here, a is the amplitude, k is the spring constant, and m is the mass. This equation shows that the maximum speed is directly proportional to the amplitude and the square root of the spring constant divided by the mass.

What is the formula for the velocity of a mass at a specific position in simple harmonic motion?

The velocity of a mass at a specific position x in simple harmonic motion can be calculated using the following formula derived from the conservation of energy:

$v=\surd \frac{k}{m}\left(a^2-x^2\right)$

Here, k is the spring constant, m is the mass, a is the amplitude, and x is the position. This equation shows that the velocity depends on the position within the oscillation and the system's parameters.

How is the total mechanical energy of a mass-spring system calculated?

The total mechanical energy (E) of a mass-spring system undergoing simple harmonic motion is the sum of its kinetic energy (KE) and elastic potential energy (PE). It can be calculated using the following formula:

$E=\frac{1}{2}k{a}^2$

Here, k is the spring constant, and a is the amplitude. This formula shows that the total mechanical energy is constant and depends only on the spring constant and the amplitude of the motion.

What happens to the energy in a mass-spring system when there is no friction?

In a mass-spring system with no friction, the total mechanical energy is conserved. This means that the energy oscillates between kinetic energy (KE) and elastic potential energy (PE) without any loss. At maximum amplitude, the energy is entirely potential, while at the equilibrium position, it is entirely kinetic. The conservation of energy can be expressed as:

$\frac{1}{2}k{a}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$

This equation shows that the total mechanical energy remains constant, and energy is continuously exchanged between kinetic and potential forms.