Question

Which of the following is necessarily true regarding frequency, angular frequency, and period of a given wave? (A) \(\quad\) The magnitude of the angular frequency is larger than the magnitude of the period. (B) \(\quad\) The product of the frequency and period is equal to the angular frequency. (C) \(\quad\) The magnitude of the angular frequency is larger than the magnitude of the frequency. (D) \(\quad\) The product of the angular frequency and period is 1

Short answer

(C) The magnitude of the angular frequency is larger than the magnitude of the frequency.

Step-by-step solution

Understand the Concepts

First, understand the definitions:- Frequency (\(f\)) is the number of oscillations per second, measured in Hertz (Hz).- Angular frequency (\(\omega\)) represents how many radians per second a wave oscillates, with the relationship \( \omega = 2\pi f \).- Period (\(T\)) is the time it takes for one complete cycle of the wave, with the relationship \( T = \frac{1}{f} \).

Analyze the Options

Analyze each option based on the relationships identified.(A) \(\quad\) The magnitude of the angular frequency is larger than the magnitude of the period.(B) \(\quad\) The product of the frequency and period is equal to the angular frequency.(C) \(\quad\) The magnitude of the angular frequency is larger than the magnitude of the frequency.(D) \(\quad\) The product of the angular frequency and period is 1.

Evaluate Option (A)

Check whether the magnitude of the angular frequency (\( \omega = 2\pi f \)) is indeed larger than the magnitude of the period (\( T = \frac{1}{f} \)).Compare:\[ \omega = 2\pi f \quad vs \quad T = \frac{1}{f} \]The angular frequency \( \omega \) depends on both \( 2\pi \) and \( f \). This relationship doesn't ensure a consistent comparison with \( T = \frac{1}{f} \), thus it may or may not be true.

Evaluate Option (B)

Check the product of the frequency and period:\[ f \cdot T = f \cdot \frac{1}{f} = 1 eq \omega \]This means the product is not equal to the angular frequency, so this option is false.

Evaluate Option (C)

Check whether the angular frequency \( \omega = 2\pi f \) is larger than the frequency \( f \).Since \( 2\pi \approx 6.28 \), it implies:\[ \omega = 2\pi f \quad > \quad f \]This option is true as \( \omega \) is always greater than \( f \) by a multiplying factor of \( 2\pi \).

Evaluate Option (D)

Check whether the product of the angular frequency and period is 1:\[ \omega \cdot T = 2\pi f \cdot \frac{1}{f} = 2\pi eq 1 \]This option is false as it results in \( 2\pi \), not 1.

Key concepts

Frequency

Frequency (\(f\)) is the measure of how many cycles or oscillations a wave completes in one second. The unit of frequency is Hertz (Hz). Think of it like the heartbeat of the wave, indicating how fast the wave is 'beating' every second.
For instance, if you have a wave with a frequency of 5 Hz, it means the wave cycles 5 times every second.
Calculating the frequency of a wave involves counting the complete cycles of the wave in a given time period and dividing by that time period. It's as simple as counting the number of ups and downs in one second!

Angular Frequency

Angular frequency (\(\omega\)) tells us how much angle the wave covers per second, measured in radians per second. The relationship between angular frequency and regular frequency is given by \(\omega = 2\pi f\). This means we multiply the normal frequency by \(2\pi\) to get the angular frequency.
Angular frequency gives a different perspective on how a wave functions. Think of it like the circular motion perspective of waves. Imagine a wave going around a circle; \(\omega\) tells you how fast it completes one full circle.
It is always a larger number than the regular frequency because \(2\pi\) is approximately 6.28. Therefore, if a wave has a frequency of 1 Hz, its angular frequency will be \(6.28\; \text{radians per second}\).

Period

The period (\(T\)) is the time it takes for one complete cycle of the wave. It is essentially the reciprocal of the frequency: \(T = \frac{1}{f}\).
To put it simply, if it takes 0.2 seconds for one complete wave cycle, then the frequency of the wave is \(\frac{1}{0.2} = 5\; \text{Hz}\). This means 5 cycles occur every second.
Understanding the period helps in visualizing the length of time each wave cycle occupies. For practical applications, it's crucial in fields like engineering and physics, where timing of wave cycles can affect outcomes and processes.
One complete wave cycle includes going from the starting point, reaching the peak, dropping to the lowest point, and coming back to the start.