Physics Problems Grade 12 With Solutions: Wave And Sound
Hey guys! Physics can be mind-bending, especially when you're diving into waves and sound. Let's tackle some Grade 12 physics problems together, complete with solutions. Buckle up, because we're about to make these concepts crystal clear!
Understanding Wave Basics
Let's start with the basics of wave. Waves are disturbances that transfer energy through a medium without transferring matter. When we talk about waves, we often discuss key characteristics like wavelength, frequency, and speed. Understanding these concepts will lay a solid foundation for tackling more complex problems.
Wavelength, Frequency, and Speed
The relationship between wavelength (λ), frequency (f), and speed (v) is fundamental: v = fλ. Wavelength is the distance between two successive crests or troughs of a wave, frequency is the number of waves passing a point per unit time, and speed is how fast the wave is propagating. Knowing how these three properties relate is crucial.
For example, imagine a wave with a frequency of 10 Hz and a wavelength of 2 meters. To find the speed, you simply multiply the frequency by the wavelength: v = 10 Hz * 2 m = 20 m/s. Easy peasy!
Types of Waves
There are two main types of waves: transverse and longitudinal. Transverse waves, like light waves, have oscillations perpendicular to the direction of energy transfer. Longitudinal waves, like sound waves, have oscillations parallel to the direction of energy transfer.
Consider a slinky. If you shake it up and down, you create a transverse wave. If you push and pull it, you create a longitudinal wave. Visualizing these different types of waves can really help when solving problems.
Sound Wave Problems
Now, let's focus on sound waves. Sound waves are longitudinal waves that travel through a medium, such as air, water, or solids. They are produced by vibrating objects, and their speed depends on the properties of the medium.
Calculating Sound Speed
The speed of sound in air is affected by temperature. A common formula to approximate the speed of sound in air is v = 331 + 0.6T, where v is the speed of sound in meters per second, and T is the temperature in degrees Celsius. This formula tells us that as temperature increases, the speed of sound also increases.
Let's say you want to find the speed of sound on a warm day when the temperature is 25°C. Using the formula, v = 331 + 0.6 * 25 = 331 + 15 = 346 m/s. So, the speed of sound on that day is approximately 346 meters per second.
Doppler Effect
The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source. It's why an ambulance siren sounds higher pitched as it approaches you and lower pitched as it moves away.
The formula for the observed frequency (f') when the source is moving is: f' = f (v ± vo) / (v ± vs), where f is the original frequency, v is the speed of sound, vo is the observer's speed, and vs is the source's speed. The signs depend on whether the source and observer are moving towards or away from each other.
For instance, imagine an ambulance with a siren emitting a frequency of 500 Hz is traveling towards you at 20 m/s. You are standing still. The speed of sound is 343 m/s. To find the observed frequency, you use the formula: f' = 500 * (343 + 0) / (343 - 20) = 500 * 343 / 323 ≈ 530.96 Hz. The siren sounds like it has a higher pitch, roughly 530.96 Hz, as the ambulance approaches.
Interference and Superposition
Waves can interfere with each other, resulting in constructive or destructive interference. When waves superpose, the resulting amplitude is the sum of the amplitudes of the individual waves.
Constructive Interference
Constructive interference occurs when waves are in phase, meaning their crests and troughs align. This results in a wave with a larger amplitude. For example, if two waves with amplitudes of 1 meter each interfere constructively, the resulting wave will have an amplitude of 2 meters.
Destructive Interference
Destructive interference occurs when waves are out of phase, meaning the crest of one wave aligns with the trough of another. This results in a wave with a smaller amplitude, or even complete cancellation if the amplitudes are equal. If two waves with amplitudes of 1 meter each interfere destructively, the resulting wave will have an amplitude of 0 meters.
Resonance
Resonance occurs when an object is driven by a force that oscillates at its natural frequency. This causes the object to vibrate with a large amplitude. A classic example is pushing a child on a swing. If you push at the right frequency, the swing will go higher and higher.
Natural Frequency
Every object has a natural frequency at which it vibrates most easily. For a string fixed at both ends, the natural frequencies are given by fn = n(v / 2L), where n is an integer (1, 2, 3, ...), v is the speed of the wave on the string, and L is the length of the string. These frequencies are also known as harmonics.
For example, consider a guitar string that is 0.6 meters long, and the speed of the wave on the string is 300 m/s. The fundamental frequency (n = 1) is f1 = 1 * (300 / (2 * 0.6)) = 250 Hz. The second harmonic (n = 2) is f2 = 2 * (300 / (2 * 0.6)) = 500 Hz. Each harmonic is a multiple of the fundamental frequency.
Example Problems and Solutions
Let's solidify these concepts with some example problems.
Problem 1: Wavelength and Frequency
A sound wave has a frequency of 440 Hz and a speed of 343 m/s. What is its wavelength?
Solution:
Using the formula v = fλ, we can rearrange it to solve for wavelength: λ = v / f. Plugging in the values, λ = 343 m/s / 440 Hz ≈ 0.78 meters. So, the wavelength of the sound wave is approximately 0.78 meters.
Problem 2: Doppler Effect
A train is moving towards you at a speed of 30 m/s, and its whistle emits a sound with a frequency of 600 Hz. What frequency do you hear if the speed of sound is 343 m/s?
Solution:
Using the Doppler effect formula, f' = f (v + vo) / (v - vs). Since you are stationary, vo = 0. Thus, f' = 600 * (343 + 0) / (343 - 30) = 600 * 343 / 313 ≈ 657.51 Hz. You hear a frequency of approximately 657.51 Hz.
Problem 3: Interference
Two sound waves with the same frequency and amplitude meet at a point. If they are completely in phase, and each has an amplitude of 0.5 meters, what is the resulting amplitude?
Solution:
Since the waves are completely in phase, they interfere constructively. The resulting amplitude is the sum of the individual amplitudes: 0.5 m + 0.5 m = 1 meter. The resulting wave has an amplitude of 1 meter.
Problem 4: Resonance in a Tube
A tube that is open at both ends has a length of 1.5 meters. If the speed of sound is 343 m/s, what is the fundamental frequency of the tube?
Solution:
For a tube open at both ends, the fundamental frequency is given by f1 = v / 2L. Plugging in the values, f1 = 343 m/s / (2 * 1.5 m) ≈ 114.33 Hz. The fundamental frequency of the tube is approximately 114.33 Hz.
Tips for Solving Physics Problems
Here are some killer tips to help you ace those physics problems:
- Read the problem carefully: Understand what the problem is asking before you start plugging in numbers.
- Draw a diagram: Visualizing the problem can make it easier to understand.
- Identify the knowns and unknowns: Write down what you know and what you need to find.
- Choose the right formula: Make sure you're using the correct formula for the situation.
- Show your work: This makes it easier to track your progress and identify any mistakes.
- Check your units: Make sure your units are consistent throughout the problem.
- Think about the answer: Does your answer make sense in the context of the problem?
Conclusion
So, there you have it! We've covered a range of physics problems related to waves and sound, complete with solutions and explanations. Remember, practice makes perfect, so keep working on those problems and don't be afraid to ask for help when you need it. Physics might seem tough, but with a solid understanding of the basic principles and a bit of practice, you'll be solving problems like a pro in no time! Keep rocking, physics pals!
