Problem 79 - College Physics

Question

During an earthquake, a house plant of mass \(15.0 \mathrm{~kg}\) in a tall building oscillates with a horizontal amplitude of \(10.0 \mathrm{~cm}\) at \(0.50 \mathrm{~Hz}\). What are the magnitudes of (a) the maximum velocity, (b) the maximum acceleration, and (c) the maximum force on the plant? (Assume SHM.)

Step-by-Step Solution

Answer

(a) Maximum velocity: \(0.314 \text{ m/s}\). (b) Maximum acceleration: \(0.986 \text{ m/s}^2\). (c) Maximum force: \(14.79 \text{ N}\).

Step 1: Understand Simple Harmonic Motion (SHM)

In simple harmonic motion, the parameters of motion like maximum velocity and maximum acceleration are determined by the angular frequency (\(\omega\)) and amplitude. We need to compute \(\omega\) using the given frequency.

Step 2: Calculate Angular Frequency

The angular frequency \(\omega\) is calculated using the formula \(\omega = 2\pi f\), where \(f\) is the frequency. Here, \(f = 0.50 \text{ Hz}\). Thus, \[ \omega = 2\pi \times 0.50 = \pi \text{ radians per second} \].

Step 3: Find Maximum Velocity

The formula for maximum velocity in SHM is \(V_{max} = \omega A\), where \(A\) is the amplitude. Given, \(A = 10.0 \text{ cm} = 0.10 \text{ m}\) and \(\omega = \pi\):\[ V_{max} = \pi \times 0.10 = 0.10\pi \text{ m/s} \approx 0.314 \text{ m/s} \].

Step 4: Determine Maximum Acceleration

The formula for maximum acceleration in SHM is \(a_{max} = \omega^2 A\). Using \(\omega = \pi\) and \(A = 0.10 \text{ m}\):\[ a_{max} = \pi^2 \times 0.10 \approx 0.986 \text{ m/s}^2 \].

Step 5: Compute Maximum Force

Use Newton's second law for the maximum force applied to the plant: \(F_{max} = m \cdot a_{max}\). Given mass \(m = 15.0 \text{ kg}\):\[ F_{max} = 15.0 \times 0.986 = 14.79 \text{ N} \].

Key Concepts

Angular Frequency

Angular frequency is a vital concept when analyzing simple harmonic motion (SHM). It tells us how fast an object oscillates back and forth in a circular path. You calculate it using the formula:

• \( \omega = 2\pi f \)

where \( f \) is the frequency given in hertz (Hz). Angular frequency is measured in radians per second. It essentially links the oscillation period with the speed of the object. In our earthquake scenario, the frequency is \( 0.50 \) Hz, giving us \( \omega = \pi \) radians per second. Knowing \( \omega \) helps calculate other parameters like velocity and acceleration, making it a cornerstone for understanding motion within the context of SHM.

Maximum Velocity

The maximum velocity of an object in SHM reveals the highest speed achieved during its oscillation. This happens when the object passes through its equilibrium position. To find this, use the equation:

• \( V_{max} = \omega A \)

where \( A \) is the amplitude, the maximum distance from the equilibrium position. In our example, the amplitude is \( 10.0 \) cm or \( 0.10 \) m. With \( \omega = \pi \), the maximum velocity comes out as \( 0.10\pi \) m/s or approximately \( 0.314 \) m/s. This means the house plant reaches its fastest horizontal movement at \( 0.314 \) m/s.

Maximum Acceleration

Maximum acceleration in SHM signifies the highest rate of change of velocity. It usually happens when the object is at the extremities of its motion. The formula used to determine this acceleration is:

• \( a_{max} = \omega^2 A \)

where \( \omega^2 \) is the square of the angular frequency and \( A \) is the amplitude. With our given \( \omega = \pi \) and amplitude \( 0.10 \) m, we get \( a_{max} = \pi^2 \times 0.10 \approx 0.986 \) m/s². This value captures how quickly the velocity shifts, especially crucial during intense motion like an earthquake.

Newton's Second Law

Newton's Second Law connects force, mass, and acceleration, forming a backbone of classical mechanics. In formula terms, it is expressed as:

• \( F = m \times a \)

This principle allows us to calculate the maximum force acting on an object in SHM by multiplying its mass by the maximum acceleration. For the house plant weighing \( 15.0 \) kg and an acceleration of \( 0.986 \) m/s², the force amounts to \( 15.0 \times 0.986 = 14.79 \) N. This illustrates the maximum push or pull the plant experiences due to its oscillatory motion during the earthquake.