Question

In an electric shaver, the blade moves back and forth over a distance of \(2.00 \mathrm{~mm}\). The motion is simple harmonic, with frequency \(120 \mathrm{~Hz}\). Find \((a)\) the amplitude, \((b)\) the maximum blade speed, and ( \(c\) ) the maximum blade acceleration.

Short answer

The amplitude of the motion is \( 1.00 \times 10^{-3} \, m \), the maximum blade speed is \( 0.240 \pi \, m/s \), and the maximum acceleration is \( 57.6 \pi^2 \, m/s^2 \).

Step-by-step solution

Find the amplitude

The amplitude is the maximum displacement from the equilibrium position. Given the blade moves back and forth over a distance of 2.00mm, the amplitude is half this distance. Thus, the amplitude \( A \) is \( A = \frac{2.00 \, mm}{2} = 1.00 \, mm = 1.00 \times 10^{-3} \, m \).

Calculate the Angular Frequency

The frequency of oscillation is given. The angular frequency \( \omega \) is related to the frequency \( f \) by the equation \( \omega = 2\pi f \). Substituting the given frequency, we obtain \( \omega = 2 \pi (120 \, Hz) = 240 \pi \, rad/s \).

Determine the Maximum Blade Speed

The maximum speed occurs when the object passes through the equilibrium position and is given by \( v_{max} = \omega A \). Substituting the values of \( \omega \) and \( A \) obtained in the previous steps, \( v_{max} = 240 \pi \, rad/s \times 1.00 \times 10^{-3} \, m = 0.240 \pi \, m/s \).

Determine the Maximum Blade Acceleration

The maximum acceleration happens when the object is at its maximum displacement and is given by \( a_{max} = \omega^2 A \). Substituting the values of \( \omega \) and \( A \) obtained in the previous steps, \( a_{max} = (240 \pi \, rad/s)^2 \times 1.00 \times 10^{-3} \, m = 0.240^2 \pi^2 \, m/s^2 = 57.6 \pi^2 \, m/s^2 \).

Key concepts

Amplitude in SHM

Understanding the amplitude in simple harmonic motion (SHM) is fundamental to grasping the nature of this type of oscillation. The amplitude, symbolized as 'A', is the maximum distance an object moves from its equilibrium position. In our example, a blade in an electric shaver moves back and forth, creating an oscillation. The total distance covered by the blade is 2.00 mm, indicating that it moves 1.00 mm to one side and then 1.00 mm to the other. This distance quantification is vital as it represents the amplitude of the blade's oscillation; hence, the amplitude is 1.00 mm or 1.00 x 10^-3 m. It's crucial for students to note that the amplitude is always positive and does not depend on time, thus giving a constant value reflecting the extent of the oscillation.

When studying SHM, remember that the amplitude is the 'size' of the wave and it does not diminish over time in this ideal scenario. This makes it a key parameter in understanding the energy associated with a system undergoing SHM.

Angular Frequency

The concept of angular frequency, denoted as \( \omega \), is pivotal in the study of SHM since it relates to how often the oscillation occurs. Angular frequency is not the same as the frequency \( f \) we commonly talk about, which simply counts how many cycles occur per second. Instead, angular frequency measures the rate of change of the phase of the sinusoidal wave, taking into account the 2\( \pi \) radians in a cycle. The relationship is \( \omega = 2\pi f \).

In our exercise example, the blade's frequency is given as 120 Hz. By using the formula \( \omega = 2\pi f \), we calculate that the angular frequency is 240\( \pi \) rad/s. This angular frequency is essential for further calculations in SHM, as it's used to determine both the maximum speed and maximum acceleration of the oscillating object.

Maximum Speed in SHM

The maximum speed in SHM occurs when the object passes through the equilibrium position. It's the point where kinetic energy is highest, as all the energy of the system is kinetic. The formula to find this speed is \( v_{max} = \omega A \), where 'v_max' denotes the maximum speed, 'A' is the amplitude, and \( \omega \) is the angular frequency.

In the given exercise, by using the angular frequency of 240\( \pi \) rad/s and the amplitude of 1.00 x 10^-3 m, we determine the maximum blade speed to be 0.240\( \pi \) m/s. It's important to discern that this speed is not constant but varies throughout the cycle of the motion; it's zero at the maximum displacement and maximum at the equilibrium point.

Maximum Acceleration in SHM

Maximum acceleration in SHM is encountered when the object is at the maximum excursion from the equilibrium point—at the amplitude. This is due to the restoring force being strongest at this point, following Hooke's Law in the case of mechanical systems. To quantify this acceleration, we use the equation \( a_{max} = \omega^2 A \), with 'a_max' representing the maximum acceleration.

From the exercise, when the blade with amplitude 1.00 x 10^-3 m and angular frequency 240\( \pi \) rad/s is at its maximum displacement, the resulting maximum acceleration is calculated to be 57.6\( \pi^2 \) m/s^2. Understanding the relationship between acceleration and displacement in SHM is key to recognizing the repetitive and predictable nature of such motions, highlighted by the fact that this acceleration, like the maximum speed, is not constant but also varies during the oscillation.