Intro to Simple Harmonic Motion (Horizontal Springs): Videos & Practice Problems

Simple harmonic motion (SHM) is exemplified by a mass-spring system, where the force is described by Hooke's law, F = kx. Key concepts include amplitude, period, frequency, and angular frequency, with relationships such as f = 1/T and ω = km. Velocity and acceleration vary sinusoidally, with maximum values at amplitude and equilibrium, essential for understanding oscillatory motion.

Intro to Simple Harmonic Motion

Simple harmonic motion (SHM) is a fundamental concept in physics, often exemplified by the mass-spring system. In this system, when a mass is attached to a spring and displaced from its equilibrium position, it experiences a restoring force described by Hooke's Law, which states that the force exerted by the spring is proportional to the displacement from equilibrium: \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement.

As the mass is pulled and released, it oscillates back and forth around the equilibrium position. The maximum displacement from this position is known as the amplitude, denoted as \( A \), which can be positive or negative, indicating the direction of displacement. At the amplitude points, the velocity of the mass is zero, while it reaches maximum velocity as it passes through the equilibrium position.

The motion can be characterized by two key variables: the period \( T \) and the frequency \( f \). The period is the time taken to complete one full cycle of motion, while the frequency is the number of cycles per second, given by the relationship \( f = \frac{1}{T} \). For instance, if the period is 2 seconds, the frequency would be 0.5 Hz. Additionally, the angular frequency \( \omega \) is defined as \( \omega = 2\pi f \), with units in radians per second.

In terms of forces and acceleration, the maximum force occurs at the maximum displacement, where the spring force is greatest. Conversely, at the equilibrium position, the force is zero. The relationship between force and acceleration is governed by Newton's second law, \( F = ma \). Thus, maximum force corresponds to maximum acceleration at the endpoints of the motion, while both force and acceleration are zero at the equilibrium position. Interestingly, the velocity is at its maximum when both force and acceleration are zero, highlighting the unique interplay between these variables in simple harmonic motion.

Example

In a mass-spring system, when a mass is displaced from its equilibrium position, it undergoes simple harmonic motion. In this example, a mass is pulled 1 meter away from its equilibrium and released from rest. The maximum displacement on either side of the equilibrium is known as the amplitude, which in this case is 1 meter.

Next, we consider the period of the motion, denoted as \( T \). The period is the time it takes to complete one full cycle of motion. Here, it is given that the mass takes 2 seconds to reach the maximum displacement on the opposite side. However, this time only accounts for half of the period. Therefore, to find the full period, we double this time, resulting in a total period of \( T = 4 \) seconds.

To relate the angular frequency \( \omega \) to the period, we can use the formula:

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

Substituting the known period into the equation gives:

\( \omega = \frac{2\pi}{4} = \frac{\pi}{2} \approx 1.57 \, \text{radians per second} \)

This angular frequency indicates how quickly the mass oscillates back and forth in its motion. Understanding these concepts—amplitude, period, and angular frequency—provides a solid foundation for analyzing simple harmonic motion in mass-spring systems.

A mass-spring system with an angular frequency ω = 8π rad/s oscillates back and forth. (a) Assuming it starts from rest, how much time passes before the mass has a speed of 0 again? (b) How many full cycles does the system complete in 60s?

• A. 1/4 s; 480 cycles
• B. 1/8 s; 480 cycles
• C. 1/8 s; 240 cycles
• D. 1/4 s; 240 cycles

Answer: (C)

Equations of Simple Harmonic Motion

In the study of mass-spring systems and simple harmonic motion, understanding the relationship between position, velocity, and acceleration is crucial. As a mass oscillates on a spring, its acceleration varies, making it necessary to use time-dependent equations rather than the traditional kinematic equations. The key equations for simple harmonic motion can be expressed in terms of time, incorporating angular frequency (ω) and amplitude (A).

The fundamental equations for position (x), velocity (v), and acceleration (a) in simple harmonic motion are:

1. Position: x(t) = A \cos(\omega t)

2. Velocity: v(t) = -A \omega \sin(\omega t)

3. Acceleration: a(t) = -A \omega^2 \cos(\omega t)

Here, ω is the angular frequency, which can be calculated using the formula:

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

where k is the spring constant and m is the mass. This relationship is vital as it connects the linear frequency and the period of oscillation.

To determine the maximum values of these quantities, we note that the maximum displacement occurs at the amplitude (A), the maximum velocity is given by \(v_{max} = A \omega\), and the maximum acceleration is \(a_{max} = A \omega^2\). The sine and cosine functions oscillate between -1 and 1, meaning that the maximum values occur when these functions reach their extremities.

For practical applications, consider a mass of 4 kg attached to a spring with a spring constant (k). If the mass is pulled to an amplitude of 2 meters, the angular frequency can be calculated. For example, if k is known, ω can be derived, and subsequently, the velocity at a specific time (t) can be calculated using the velocity function. If t = 0.5 seconds, the velocity can be computed as:

\(v(0.5) = -A \omega \sin(\omega \cdot 0.5)\)

Additionally, if the position is given, the acceleration can be found using the position-dependent formula:

\(a(x) = -\frac{k}{m} x\)

Finally, the period of oscillation (T) can be determined using the relationship between angular frequency and period:

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

By mastering these equations and their applications, one can effectively analyze and predict the behavior of mass-spring systems in simple harmonic motion.

A 4-kg mass on a spring is released 5 m away from equilibrium position and takes 1.5 s to reach its equilibrium position. (a) Find the spring's force constant. (b) Find the object's max speed.

• A. 3.28 N/m; 4.52 m/s
• B. 4.39 N/m; 5.25 m/s
• C. 17.54 N/m; 10.47 m/s
• D. 2.01 N/m; 3.54 m/s

Answer: (B)

Example

In a mass-spring system, a 4-kilogram mass oscillates at a frequency of 2 hertz, achieving a maximum speed of 10 meters per second as it crosses the equilibrium position. This maximum speed, denoted as \( v_{\text{max}} \), occurs at the equilibrium point, where the kinetic energy is at its peak. To determine the time taken to move from the equilibrium position to the maximum displacement (amplitude), we recognize that this interval represents one-quarter of the oscillation period.

The period \( T \) of oscillation is the inverse of frequency \( f \), expressed as:

\[ T = \frac{1}{f} \]

Given that \( f = 2 \, \text{Hz} \), the period is:

\[ T = \frac{1}{2} \, \text{s} \]

Thus, the time to reach maximum displacement is:

\[ t = \frac{T}{4} = \frac{1/2}{4} = \frac{1}{8} \, \text{s} \]

Next, to find the amplitude \( A \), we utilize the relationship between maximum speed and amplitude, given by:

\[ v_{\text{max}} = A \cdot \omega \]

Where \( \omega \) is the angular frequency, calculated as:

\[ \omega = 2 \pi f \]

Substituting the frequency:

\[ \omega = 2 \pi \cdot 2 = 4 \pi \, \text{rad/s} \]

Now, rearranging the equation for amplitude gives:

\[ A = \frac{v_{\text{max}}}{\omega} = \frac{10}{4 \pi} = \frac{5}{2 \pi} \, \text{m} \]

Finally, to find the maximum acceleration \( a_{\text{max}} \), we use the formula:

\[ a_{\text{max}} = A \cdot \omega^2 \]

Substituting the values for amplitude and angular frequency:

\[ a_{\text{max}} = \left(\frac{5}{2 \pi}\right) \cdot (4 \pi)^2 \]

Calculating \( \omega^2 \):

\[ (4 \pi)^2 = 16 \pi^2 \]

Thus, the maximum acceleration becomes:

\[ a_{\text{max}} = \frac{5}{2 \pi} \cdot 16 \pi^2 = 5 \cdot 8 \pi = 40 \pi \, \text{m/s}^2 \]

In summary, the time to reach maximum displacement is \( \frac{1}{8} \, \text{s} \), the amplitude is \( \frac{5}{2 \pi} \, \text{m} \), and the maximum acceleration is \( 40 \pi \, \text{m/s}^2 \).

What is the equation for the position of a mass moving on the end of a spring which is stretched 8.8cm from equilibrium and then released from rest, and whose period is 0.66s? What will be the object's position after 1.4s?

• A. .88cos(19.04t); .79 m
• B. .088cos(19.04t); .08 m
• C. .88cos(9.52t); .86 m
• D. .088cos(9.52t); .064 m

Answer: (D)

Example

In this example, we explore the relationship between the velocity of a particle and its frequency of motion, using the given velocity equation:

Velocity Equation: \( v(t) = -6 \sin(3\pi t) \

To find the frequency of motion, we first identify the angular frequency (\( \omega \)) from the equation. The general form of the velocity equation is:

General Form: \( v(t) = -A \omega \sin(\omega t) \

From our equation, we can see that:

• Amplitude (\( A \)) is related to the coefficient of the sine function, which is \( 6 \).
• Angular frequency (\( \omega \)) is \( 3\pi \).

Using the relationship between angular frequency and frequency, we have:

Frequency Equation: \( \omega = 2\pi f \

Substituting the value of \( \omega \):

\( 3\pi = 2\pi f \

Solving for frequency (\( f \)) gives:

\( f = \frac{3\pi}{2\pi} = \frac{3}{2} \) Hz.

Next, we determine the amplitude. We know from our earlier analysis that:

Amplitude Relationship: \( A \omega = 6 \

Substituting \( \omega = 3\pi \):

\( A (3\pi) = 6 \

Thus, the amplitude (\( A \)) can be calculated as:

\( A = \frac{6}{3\pi} = \frac{2}{\pi} \).

Finally, to find the velocity at a specific time, we substitute \( t = 0.5 \) seconds into the velocity equation:

\( v(0.5) = -6 \sin(3\pi \cdot 0.5) \

Calculating this gives:

\( v(0.5) = -6 \sin(\frac{3\pi}{2}) = -6 \cdot (-1) = 6 \) m/s.

In summary, we have determined that the frequency of motion is \( \frac{3}{2} \) Hz, the amplitude is \( \frac{2}{\pi} \), and the velocity at \( t = 0.5 \) seconds is \( 6 \) m/s.

Here’s what students ask on this topic:

What is simple harmonic motion in a mass-spring system?

Simple harmonic motion (SHM) in a mass-spring system occurs when a mass attached to a spring oscillates back and forth around an equilibrium position. The restoring force is described by Hooke's law, $F=kx$, where $k$ is the spring constant and $x$ is the displacement from equilibrium. Key parameters include amplitude (maximum displacement), period (time for one complete cycle), frequency (cycles per second), and angular frequency (related to frequency by $\omega =2\pi f$). The motion is sinusoidal, with velocity and acceleration varying periodically.

How do you calculate the period and frequency of a mass-spring system?

The period $T$ of a mass-spring system is the time it takes to complete one full oscillation. It is given by $T=\frac{2}{\pi }\sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant. The frequency $f$ is the number of oscillations per second and is the reciprocal of the period: $f=\frac{1}{T}$. The units of frequency are hertz (Hz).

What is the relationship between angular frequency and the spring constant in SHM?

In simple harmonic motion (SHM), the angular frequency $\omega$ is related to the spring constant $k$ and the mass $m$ by the equation $\omega =\sqrt{\frac{k}{m}}$. This relationship shows that the angular frequency increases with a stiffer spring (higher $k$) and decreases with a larger mass. Angular frequency is measured in radians per second (rad/s).

How do you determine the maximum velocity and acceleration in a mass-spring system?

The maximum velocity $v$_max in a mass-spring system occurs at the equilibrium position and is given by $v$_max=Aω, where $A$ is the amplitude and $\omega$ is the angular frequency. The maximum acceleration $a$_max occurs at the maximum displacement (amplitude) and is given by $a$_max=Aω². These relationships highlight the sinusoidal nature of SHM, where velocity and acceleration vary periodically.

What are the key differences between position, velocity, and acceleration equations in SHM?

In simple harmonic motion (SHM), the position, velocity, and acceleration equations are sinusoidal functions of time. The position $x$ is given by $x=A\cos \left(\omega t\right)$. The velocity $v$ is the derivative of position and is given by $v=-A\omega \sin \left(\omega t\right)$. The acceleration $a$ is the derivative of velocity and is given by $a=-A\omega 2\cos \left(\omega t\right)$. These equations show how position, velocity, and acceleration are phase-shifted relative to each other.