Problem 13 - University Physics with Modern Physics

Question

The point of the needle of a sewing machine moves in SHM along the \(x\) -axis with a frequency of 2.5 Hz. At \(t=0\) its position and velocity components are \(+1.1 \mathrm{cm}\) and \(-15 \mathrm{cm} / \mathrm{s},\) respectively. (a) Find the acceleration component of the needle at \(t=0\) . (b) Write equations giving the position, velocity, and acceleration components of the point as a function of time.

Step-by-Step Solution

Answer

The acceleration at \(t=0\) is approximately \[-273.9\text{ cm/s}^2.\]

Step 1: Understand Simple Harmonic Motion (SHM)

In SHM, the motion is sinusoidal in time. The position, velocity, and acceleration can be expressed as: \[x(t) = A \cos(\omega t + \phi)\]\[v(t) = -A\omega \sin(\omega t + \phi)\]\[a(t) = -A\omega^2 \cos(\omega t + \phi)\]where \(A\) is amplitude, \(\omega\) is angular frequency, and \(\phi\) is the phase.

Step 2: Determine Angular Frequency

The frequency given is \(2.5\text{ Hz}\). The angular frequency \(\omega\) is calculated using \(\omega = 2\pi f\):\[\omega = 2\pi \times 2.5 = 5\pi \text{ rad/s}\]

Step 3: Calculate Initial Conditions

At \(t=0\), position \(x(0) = 1.1 \text{ cm}\), velocity \(v(0) = -15 \text{ cm/s}\). Use these to find amplitude \(A\) and phase \(\phi\).

Step 4: Find Phase \(\phi\) Using Position

From \(x(0) = A \cos \phi\) and given that \(x(0) = 1.1\text{ cm}\), let \(A = \text{Variable}\). We will first use velocity equation to find \(A\).

Step 5: Use Velocity to Find Amplitude and Phase

From\[v(0) = -A\omega\sin(\phi) = -15\text{ cm/s}\]Plug in \(\omega = 5\pi\) and solve:\[-A \times 5\pi \sin(\phi) = -15\]Solve for \(A\sin(\phi)\):\[A \sin(\phi) = \frac{15}{5\pi}\]

Step 6: Equate Position and Velocity

Express \(\cos(\phi)\) in terms of known quantities:\[A \cos(\phi) = 1.1\]Now solve for \(A\) and \(\phi\) by squaring and adding the equations:\[A^2(\cos^2(\phi) + \sin^2(\phi)) = 1.1^2 + \left(\frac{15}{5\pi}\right)^2\]

Step 7: Determine Amplitude \(A\)

By simplifying the above:\[A^2 = 1.21 + \left(\frac{15}{5\pi}\right)^2\approx 1.21 + 0.227^2 = 1.21 + 0.0515\]\[A = \sqrt{1.2615} \approx 1.122 cm\]

Step 8: Calculate Phase \(\phi\)

Using \(A\cos(\phi) = 1.1\text{ cm}\):\[\cos(\phi) = \frac{1.1}{1.122}\]\[\cos(\phi) \approx 0.980\]\(\phi \approx \arccos(0.980)\approx 0.2 ext{ radians}\)

Step 9: Find Acceleration at t=0

Use acceleration formula:\[a(0) = -A\omega^2 \cos(\phi)\]\[a(0) = -(1.122) \times (5\pi)^2 \times 0.980\]\[a(0) \approx -273.9\text{ cm/s}^2\]

Step 10: Write Position, Velocity, and Acceleration as Functions of Time

The position, velocity, and acceleration are:\[x(t) = 1.122\cos(5\pi t + 0.2)\]\[v(t) = -1.122 \times 5\pi \sin(5\pi t + 0.2)\]\[a(t) = -1.122 \times (5\pi)^2 \cos(5\pi t + 0.2)\]

Key Concepts

Angular Frequency

When dealing with Simple Harmonic Motion (SHM), understanding angular frequency is crucial. This concept describes how quickly an object moves through its cycle in SHM. Essentially, it tells us how fast the oscillations occur.

Angular frequency is denoted by the symbol \( \omega \) and is related to the regular frequency \( f \), which is the number of cycles per second. The relationship is defined by the formula:

• \( \omega = 2\pi f \)

In this case, since the frequency is given as 2.5 Hz, using the formula we calculate:

• \( \omega = 2\pi \times 2.5 = 5\pi \) rad/s

You might visualize angular frequency like how fast the hands on a clock move. The faster they move, the larger the angular frequency. This aspect is fundamental in describing the motion dynamics of any object undergoing SHM.

Amplitude

Amplitude in Simple Harmonic Motion (SHM) expresses the maximum extent of displacement from the equilibrium position.

In simpler terms, amplitude is how far the object swings either side of its rest position. It's denoted by the letter \( A \) and is always a positive value because it represents a distance.

In our exercise, after solving the equations, we found that the amplitude \( A \) of the sewing needle's motion is approximately \( 1.122 \) cm. Let's explore how to determine this:

• Using the position equation, \( x(0) = A \cos(\phi) \) and the velocity equation, \( v(0) = -A\omega \sin(\phi) \), we could solve for both \( A \) and the phase angle \( \phi \).
• For both the position and the velocity equations, we derived expressions involving \( \sin(\phi) \) and \( \cos(\phi) \), which were squared and summed to attain \( A^2 \).
• Simplifying these results yielded \( A = \sqrt{1.2615} \approx 1.122 \) cm.

Amplitude is an important concept as it helps define the scale of oscillation in SHM, affecting how large or small the motion seems.

Phase Angle

The phase angle, often denoted by \( \phi \), is a vital component in understanding SHM, as it details the specific starting point of motion within its cycle.

It can seem a little abstract, as it doesn’t directly describe a physical property, but rather indicates the initial angle at \( t=0 \). Essentially, it helps us know whether the motion starts at the maximum displacement, the midpoint, or somewhere in between.

• In our exercise, we first computed \( \phi \) using the known position and amplitude: \( \phi \approx \arccos(0.980) \approx 0.2 \text{ radians} \).

The phase angle makes sure that the resulting motion equations capture the initial conditions accurately, such as the given starting position and velocity.

This concept not only pinpoints the initial conditions but also helps in predicting future positions, velocities, and accelerations of objects exhibiting SHM.