Problem 3 - University Physics with Modern Physics

Question

14.3 A mass that can oscillate without friction on a horizontal surface is attached to a horizontal spring that is pulled to the right \(10.0 \mathrm{~cm}\) and is released from rest. The period of oscillation for the mass is \(5.60 \mathrm{~s}\). What is the speed of the mass at \(t=2.50 \mathrm{~s} ?\) a) \(-2.61 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\) b) \(-3.71 \cdot 10^{-2} \mathrm{~m} / \mathrm{s}\) c) \(-3.71 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\) d) \(-2.01 \cdot 10^{-1} \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

a) \(-2.61 \cdot 10^{-1}\ \text{m/s}\) b) \(3.76 \cdot 10^{-1}\ \text{m/s}\) c) \(4.43 \cdot 10^{-1}\ \text{m/s}\) d) \(6.28 \cdot 10^{-1}\ \text{m/s}\) Answer: a) \(-2.61 \cdot 10^{-1}\ \text{m/s}\)

Step 1: 1. Find the angular frequency

The period of oscillation is related to angular frequency \(\omega\) by the formula: \(\omega = \frac{2 \pi}{T}\), where \(T\) is the period. In this case, we are given the period \(T = 5.60\ \text{s}\). Thus, the angular frequency is: \(\omega = \frac{2 \pi}{5.60} \approx 1.12\ \text{rad/s}\).

Step 2: 2. Calculate the maximum speed

The maximum speed, \(v_{max}\), can be found using the equation: \(v_{max} = A\omega\), where \(A\) is the amplitude (maximum displacement) and \(\omega\) is the angular frequency. In this case, the maximum displacement is \(10.0\ \text{cm} = 0.100\ \text{m}\). So the maximum speed is: \(v_{max} = 0.100 \times 1.12 \approx 0.112\ \text{m/s}\).

Step 3: 3. Find the phase constant

Since the mass is released from rest at the maximum displacement, the phase constant \(\varphi\) will be zero.

Step 4: 4. Calculate the velocity at t = 2.50 s

We can use the equation for the velocity of an oscillating mass: \(v(t) = v_{max}\cos(\omega t + \varphi)\). Plugging in the values we found earlier: \(v(2.50) = 0.112 \cdot \cos(1.12 \times 2.50) \approx -0.262\ \text{m/s}\). Comparing this result to the provided choices, we find that the closest one is: a) \(-2.61 \cdot 10^{-1}\ \text{m/s}\). So the correct answer is option a.

Key Concepts

Angular Frequency

The concept of angular frequency is fundamental in understanding oscillatory motion, such as that of a mass attached to a spring. In physics, angular frequency (\r\text{\(\omega\)})\r) is a measure of how quickly an object traverses through an angle. It is the rate at which an object rotates or revolves relative to a central point or axis. For an oscillating system, it defines how many cycles of oscillation occur per unit time.

\rFor oscillatory motion, the angular frequency is calculated by the formula \r\text{\(\omega = \frac{2 \pi}{T}\)}\r, where \r\text{\(T\)}\r is the period of one full cycle of oscillation. The period is the time it takes to complete one cycle. Thus, a shorter period implies a higher angular frequency and more cycles per second. Higher angular frequency results in a faster oscillation.

\rIn the example problem provided, the period is 5.60 s, which gives us an angular frequency of approximately 1.12 rad/s, showing the object completes just over one radian of its cycle every second.

Amplitude in Oscillations

Amplitude in oscillations refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium. Amplitude (\r\text{\(A\)}\r) is an important characteristic of waves and oscillations, as it indicates the maximum displacement from the rest position.

\rWhen dealing with mechanical oscillators such as springs, the amplitude is the furthest distance that the mass moves from its equilibrium position. In the step by step solution, the given amplitude of 10.0 cm (converted to 0.100 m) is the distance the mass was pulled back before being released. This initial displacement represents the amplitude because the mass can oscillate this maximum distance from the equilibrium in either direction.

\rMoreover, the amplitude is directly proportional to the energy stored in an oscillating system; a larger amplitude means that more energy is stored in the system. It is also crucial to determining the maximum speed of the mass in an oscillating system, as the equation \r\text{\(v_{max} = A\omega\)}\r demonstrates. Hence, understanding amplitude is key to comprehending the energy and motion characteristics of an oscillator.

Phase Constant

The phase constant (\r\text{\(\varphi\)}\r) in oscillation is a term that accounts for the initial conditions of the system. It represents the phase of the oscillation at time equals zero and determines at what point in its cycle the oscillation begins.

\rFor instance, if you release a mass from rest at its maximum displacement from the equilibrium position, like in the problem we're examining, the phase constant is zero. This is because the oscillation starts from the maximum amplitude, which corresponds to the cosine of zero in the cosine function. The phase constant can be thought of as the 'starting angle' for the cosine wave that represents the oscillation.

\rMathematically, the phase constant comes into play in the equations of motion for oscillating systems, specifically in the velocity equation \r\text{\(v(t) = v_{max}\cos(\omega t + \varphi)\)}\r. In our example, because the object starts from the furthest point to the right and begins its motion to the left, the phase constant is indeed zero, simplifying the velocity equation and highlighting the phase constant's role in determining the motion based on initial conditions.