Question

Which of the following transverse waves has the greatest power? a) a wave with velocity \(v\), amplitude \(A\), and frequency \(f\) b) a wave of velocity \(v\), amplitude \(2 A\), and frequency \(f / 2\) c) a wave of velocity \(2 v\), amplitude \(A / 2\), and frequency \(f\) d) a wave of velocity \(2 v\), amplitude \(A\), and frequency \(f / 2\) e) a wave of velocity \(v\), amplitude \(A / 2\), and frequency \(2 f\)

Short answer

a) Wave with amplitude A, velocity v, and frequency f. b) Wave with amplitude 2A, velocity v, and frequency f/2. c) Wave with amplitude A/2, velocity 2v, and frequency f. d) Wave with amplitude A, velocity 2v, and frequency f/2. e) Wave with amplitude A/2, velocity v, and frequency 2f. Answer: Both b) and e), as each has the same power: \(B \cdot f^2 \cdot A^2 \cdot v\).

Step-by-step solution

Calculate the power of wave a

Using the formula \(P = \frac{1}{2} \cdot B \cdot f^2 \cdot A^2 \cdot v\), we can rewrite the power of wave a in terms of \(B\), \(f\), \(A\), and \(v\): \(P_a = \frac{1}{2} \cdot B \cdot f^2 \cdot A^2 \cdot v\).

Calculate the power of wave b

Using the formula for power, we can rewrite the power of wave b in terms of \(B\), \(f\), \(A\), and \(v\): \(P_b = \frac{1}{2} \cdot B \cdot (f / 2)^2 \cdot (2 A)^2 \cdot v = \frac{1}{2} \cdot B \cdot (1/4) \cdot f^2 \cdot (4 \cdot A^2) \cdot v = B \cdot f^2 \cdot A^2 \cdot v\).

Calculate the power of wave c

Using the formula for power, we can rewrite the power of wave c in terms of \(B\), \(f\), \(A\), and \(v\): \(P_c = \frac{1}{2} \cdot B \cdot f^2 \cdot (A / 2)^2 \cdot 2 v = \frac{1}{2} \cdot B \cdot f^2 \cdot (1/4 \cdot A^2) \cdot 2 v = \frac{1}{4}B \cdot f^2 \cdot A^2 \cdot v\).

Calculate the power of wave d

Using the formula for power, we can rewrite the power of wave d in terms of \(B\), \(f\), \(A\), and \(v\): \(P_d = \frac{1}{2} \cdot B \cdot (f/2)^2 \cdot A^2 \cdot 2 v = \frac{1}{2} \cdot B \cdot (1/4) \cdot f^2 \cdot A^2 \cdot 2 v = \frac{1}{4}B \cdot f^2 \cdot A^2 \cdot v\).

Calculate the power of wave e

Using the formula for power, we can rewrite the power of wave e in terms of \(B\), \(f\), \(A\), and \(v\): \(P_e = \frac{1}{2} \cdot B \cdot (2f)^2 \cdot (A/2)^2 \cdot v = \frac{1}{2} \cdot B \cdot 4 \cdot f^2 \cdot (1/4 \cdot A^2) \cdot v = B \cdot f^2 \cdot A^2 \cdot v\).

Compare the powers

Comparing the powers of all the waves, we find that: - \(P_a = \frac{1}{2}B \cdot f^2 \cdot A^2 \cdot v\) - \(P_b = B \cdot f^2 \cdot A^2 \cdot v\) - \(P_c = \frac{1}{4}B \cdot f^2 \cdot A^2 \cdot v\) - \(P_d = \frac{1}{4}B \cdot f^2 \cdot A^2 \cdot v\) - \(P_e = B \cdot f^2 \cdot A^2 \cdot v\) Clearly, the greatest power is shared by wave b and wave e, both having power equal to \(B \cdot f^2 \cdot A^2 \cdot v\). So, the correct answer is both b) and e).

Key concepts

Wave Velocity

Wave velocity, commonly symbolized as v, is a fundamental concept in wave physics, referring to the speed at which waves propagate through a medium. This medium could be anything from air and water to the vacuum of space in the case of electromagnetic waves. The velocity of a wave is determined by the type of wave and the properties of the medium, and it remains constant as long as those properties don't change.

The velocity of a transverse wave, which is the type of wave in our exercise, can be calculated using the formula:
\[ v = f\times \lambda \]
where f represents the frequency of the wave and \lambda (lambda) is the wavelength. Note that this formula implies that any changes in frequency or wavelength directly affect the wave's velocity, as long as the medium remains the same. Understanding wave velocity is essential when comparing the power of waves, as a higher velocity can contribute to a greater transfer of energy over a distance per unit time.

Wave Amplitude

Wave amplitude, denoted typically by A, is a measure of the maximum displacement of particles from their rest position as a wave passes through a medium. Think of it as the height of a wave crest above the rest level or the depth of a trough below it. The amplitude of a wave is directly related to the energy carried by the wave; a greater amplitude means the wave has more energy.

When addressing the power of waves, amplitude plays a crucial role. The formula used in the example exercise incorporates amplitude squared (\(A^2\)), meaning that the power transmitted by the wave increases with the square of the amplitude. This relationship highlights why a wave with twice the amplitude doesn't just double the power; it actually increases it by four times, provided other factors remain constant.

Wave Frequency

The frequency of a wave, symbolized as f, describes the number of cycles that pass a point in one second. Measured in Hertz (Hz), frequency is intrinsic to a wave's identity, like pitch is to a sound. High-frequency waves have closely packed cycles, while low-frequency waves have cycles that are spread out.

In the context of the transverse wave power calculation, frequency influences the energy transfer rate. According to the wave power formula used in our problem, the power is proportional to the square of the frequency (\(f^2\)). Therefore, if the frequency of a wave is doubled, all else being equal, the power transmitted by the wave will increase by a factor of four, indicating a strong dependence on frequency for the wave's power.

Power Calculation of Waves

The power transmitted by a wave is a measure of the energy transferred by the wave per unit time. In the given exercise, the power calculation for a transverse wave involves the formula:
\[P = \frac{1}{2} \cdot B \cdot f^2 \cdot A^2 \cdot v\]
In this equation, P stands for power, B for a proportionality constant that depends on the medium and type of wave, f for the wave's frequency, A for amplitude, and v for velocity. From the solution steps, we see that both square of the amplitude (\(A^2\)) and square of the frequency (\(f^2\)) have a significant impact on the power. Therefore, any changes to the amplitude or frequency will exponentially affect the wave's power.

By comparing the power calculations of various waves with different velocities, amplitudes, and frequencies, we can determine which wave carries the most energy. It's also crucial to note that the velocity and amplitude in the exercise refer to peak or maximum values, which directly correlate to the wave's energy-carrying capacity.