Question

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

Short answer

a) A wave with velocity \(v\), amplitude \(A\), and frequency \(f\) b) A wave with velocity \(v\), amplitude \(2A\), and frequency \(f/2\) c) A wave with velocity \(2v\), amplitude \(A/2\), and frequency \(f\) d) A wave with velocity \(2v\), amplitude \(A\), and frequency \(f/2\) e) A wave with velocity \(2v\), amplitude \(A/2\), and frequency \(2f\) Answer: e) A wave with velocity \(2v\), amplitude \(A/2\), and frequency \(2f\)

Step-by-step solution

Wave (a)

First, let's consider wave (a). We are given the following values: - velocity, \(v\) - amplitude, \(A\) - frequency, \(f\) We can compute the product for wave (a) as follows: \((vf^2A^2)_a = v \cdot f^2 \cdot A^2\)

Wave (b)

Now, let's consider wave (b). We are given the following values: - velocity, \(v\) - amplitude, \(2A\) - frequency, \(f/2\) We can compute the product for wave (b) as follows: \((vf^2A^2)_b = v \cdot (\frac{f}{2})^2 \cdot (2A)^2 = v \cdot \frac{f^2}{4} \cdot 4A^2 = vf^2A^2\)

Wave (c)

Next, let's look at wave (c). We are given the following values: - velocity, \(2v\) - amplitude, \(A/2\) - frequency, \(f\) We can compute the product for wave (c) as follows: \((vf^2A^2)_c = (2v) \cdot f^2 \cdot (\frac{A}{2})^2 = 2vf^2 \cdot \frac{A^2}{4} = \frac{1}{2}vf^2A^2\)

Wave (d)

Now, let's consider wave (d). We are given the following values: - velocity, \(2v\) - amplitude, \(A\) - frequency, \(f/2\) We can compute the product for wave (d) as follows: \((vf^2A^2)_d = (2v) \cdot (\frac{f}{2})^2 \cdot A^2 = 2v \cdot \frac{f^2}{4} \cdot A^2 = vf^2A^2\)

Wave (e)

Lastly, let's look at wave (e). We are given the following values: - velocity, \(2v\) - amplitude, \(A/2\) - frequency, \(2f\) We can compute the product for wave (e) as follows: \((vf^2A^2)_e = (2v) \cdot (2f)^2 \cdot (\frac{A}{2})^2 = 2v \cdot 4f^2 \cdot \frac{A^2}{4} = 2vf^2A^2\)

Conclusion

Now, let's compare the products for the different waves: - Wave (a): \((vf^2A^2)_a = vf^2A^2\) - Wave (b): \((vf^2A^2)_b = vf^2A^2\) - Wave (c): \((vf^2A^2)_c = \frac{1}{2}vf^2A^2\) - Wave (d): \((vf^2A^2)_d = vf^2A^2\) - Wave (e): \((vf^2A^2)_e = 2vf^2A^2\) From the above comparison, we can see that wave (e) has the greatest product and therefore the greatest power. So, the correct answer is option e) a wave with velocity \(2v\), amplitude \(A / 2,\) and frequency \(2f\).

Key concepts

Wave Velocity

Wave velocity, denoted by the symbol 'v', is a fundamental characteristic of a wave that describes how fast a wave travels through a medium. It is measured in units of distance per time, such as meters per second (m/s). The velocity of a wave is determined by the properties of the medium and the type of wave. For example, sound waves travel at different speeds through air, water, and solids.

When it comes to the power of a transverse wave, which is the rate at which energy is transmitted by the wave, wave velocity plays a significant role. The greater the velocity, the faster the wave can transfer energy. However, when comparing waves with different characteristics like amplitude and frequency, the overall power cannot be determined by velocity alone. The interaction of wave velocity with these other variables decides the power output.

In the textbook exercise, we find that options (c), (d), and (e) all involve waves with a velocity of double 'v'. Despite this increase, only wave (e) is considered to have the greatest power.

Wave Amplitude

Wave amplitude, represented by 'A', measures the maximum displacement of points on a wave from their rest positions. It is an indicator of the energy and intensity of the wave—the greater the amplitude, the more energy the wave carries. Amplitude is the height from the central axis (equilibrium line) to the peak (crest) or trough of the wave.

Since energy in a transverse wave is directly proportional to the square of its amplitude (that is, doubling the amplitude increases the energy by a factor of four), amplitude is crucial in determining the power of the wave. In the given exercise, waves (b) and (d) either double or maintain the amplitude of 'A', yet it's not enough to declare them the most powerful without considering frequency as well. Despite having smaller amplitude, wave (e) achieves greater power due to the combined effect with increased frequency and wave velocity.

Wave Frequency

Frequency of a wave, denoted by 'f', is defined as the number of complete wave cycles passing a point per unit of time—usually measured in cycles per second, or Hertz (Hz). Frequency is inherently linked to the energy of a wave; higher frequency means more cycles per second, which contributes to more energy transfer in that duration. Instead of thinking of it solely in terms of number of waves, consider it as a measure of how often the wave can exert force, transferring energy.

In the context of the task, energy and, consequently, power of a wave is also related to the square of the frequency, implying that if the frequency is doubled, the power increases by a factor of four. This is why in the exercise, even though wave (e) has a reduced amplitude compared to wave (a), the doubled frequency, when combined with double velocity, makes it the wave with the greatest power among the options.