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Resonances in toy balloons. Get a helium-filled balloon, Hold it near your ear and tap it. Sing into the side of it and search for resonant pitches. Blow up another balloon with air to the same diameter as the helium balloon. Tap it. Estimate the ratio of frequencies of the lowest modes (the ones you hear when you t?p) of the helium and air balloons. What frequency ratio would you predict: Compare the strength (loudness) of the resonances you get singing into the side of a helium balloon with those you get from the air balloon. Why is there such a difference?

Short Answer

Expert verified
The frequency ratio is approximately 2.82. Helium balloons are louder due to faster sound speed and less energy loss.

Step by step solution

01

Understand the Concept of Resonance

Resonance occurs when an object vibrates at a specific frequency, called the resonant frequency. In this experiment, resonances are investigated in balloons filled with different gases.
02

Analyze the Physics of Sound in Balloons

The speed of sound depends on the medium through which it travels. Helium, being lighter than air, allows sound waves to travel faster. This speed affects the pitch (frequency) of the sound you hear when tapping or singing into a balloon.
03

Calculate the Speed of Sound in Helium and Air

The speed of sound in air is approximately 343 m/s at 20°C, while in helium, it is about 965 m/s. This difference is due to the lower density and molecular weight of helium compared to air.
04

Estimate the Frequency Ratio of Lowest Modes

The fundamental frequency of a balloon's resonance is proportional to the speed of sound in the gas filling it. Thus, the frequency ratio of helium to air, \( f_{\text{helium}} / f_{\text{air}} \), is approximately \( 965 / 343 \).
05

Compare Resonance Strengths

The loudness of resonances (or the strength) also depends on how well the balloon amplifies sound, which is influenced by the speed of sound and the density of the gas. Helium balloons resonate more loudly because sound travels faster and with less energy loss than in air.
06

Conclusion

The frequency ratio of the lowest modes of the helium balloon to the air balloon is about 2.82. Helium-filled balloons provide louder resonances due to the higher speed of sound and lower energy loss.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Sound
The speed of sound is crucial in understanding how sound travels through different mediums.
In this context, we're looking at how sound moves through helium and air. Sound is basically vibrations moving through a material medium. This means the speed of sound can be affected by the type of gas it travels through.
Here's why:
  • Sound travels faster in helium than in air.
  • The speed of sound in air is typically around 343 m/s at room temperature, which is about 20°C.
  • Helium, being much less dense than air, allows sound to travel at approximately 965 m/s.
This is because helium's molecules are lighter, making them move faster and hence, transporting the sound waves more swiftly. This difference is why when you hear sound passing through a helium-filled balloon, it seems higher pitched. Understanding this concept sets the stage for exploring other aspects like frequencies and sound amplification.
Fundamental Frequency
The fundamental frequency is the lowest frequency of a periodic waveform, and in the case of balloons filled with gases, it is closely related to the speed of sound in each gas.
When you tap a balloon or sing into it, you are exciting its natural vibration modes or resonant frequencies. The fundamental frequency depends on the dimensions of the balloon and the speed of sound in the filling gas. Key points include:
  • The fundamental frequency is proportional to the speed of sound.
  • In helium-filled balloons, this frequency is higher because the sound travels faster.
  • Using the step-by-step exercise solution, for helium to air, the frequency ratio is about 965/343 or roughly 2.82.
So when you tap or sing into a helium balloon and an air balloon of the same size, you'll notice that the helium balloon has a higher fundamental frequency. This means it vibrates more quickly and, consequently, produces a higher pitch.
Sound Amplification
Sound amplification relates to how sound waves are magnified, which is observable when singing into or tapping on a balloon. Helium-filled balloons produce louder or stronger resonances compared to air-filled ones. But why is this? Here's how it works:
  • The speed of sound influences how effectively sound is amplified within the balloon.
  • Helium, with its faster-moving sound waves, leads to less energy loss as the waves travel.
  • This means the sound is not only quicker but comes across as louder or more amplified.
When sound travels through a helium-filled balloon, the vibrations are retained more effectively, contributing to a brighter, more pronounced resonance. This is why you might notice a more vibrant sound when interacting with helium balloons versus those filled with air.
Gas Density
Gas density plays a vital role in the way sound behaves in balloons. It's the density of the gas that determines how sound waves are propagated and perceived. The exercise highlighted the distinction between helium and air in terms of density, leading to different sound characteristics. Consider this:
  • Helium has a lower density than air, since its molecules are lighter.
  • This lower density allows for faster sound wave propagation.
  • Air, being denser, restricts sound waves more than helium, leading to slower propagation and lower pitch.
Practically, this means that in a helium balloon, sound can move freely and swiftly, creating a sound that's higher pitched and more intense. On the other hand, air-filled balloons dampen the sound more due to the higher density, resulting in a muffled and lower-pitched resonance. This is a key insight into why different gases produce different acoustic profiles in similar sized balloons.

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Most popular questions from this chapter

Suppose that a point \(a\) on a string at \(z_{n}=10 \mathrm{~cm}\) oscillates in harmonic motion at frequency 10 eps with amplitude \(1 \mathrm{~cm}\). Its phase is such that at \(t=0\) the point on the string is passing through its equilibrium position with upward velocity (positive displacement is upward). (a) What is the magnitude and direction of the velocity of point \(a\) at \(t=0.05 \mathrm{sec}\) ? Suppose the string parameters (mass per unit length and tension) are such that the wave velocity is \(100 \mathrm{~cm} / \mathrm{sec}\) (b) What is the wavelength of a traveling wave? What is the wavelength of a standing wave? (c) Another point \(b\) at \(z_{b}=15\) cm oscillates with the same amplitude as that at \(z_{a}=10 \mathrm{~cm}\), but with a relative phase of 180 deg with respect to the oscillation at \(\hat{ }_{20}\) Can you tell whether we have here a pure traveling wave, a pure standing wave, or a combination? (d) A third point \(c\) at \(12.5 \mathrm{~cm}\) also oscillates with the same amplitude as that at \(z_{n}\) but 180 deg out of phase with point \(a\). Point \(b\) oscillates as given above. Now tell us whether the wave is a traveling or a standing wave (or a combination).

Measuring the relative phase at the two ends of an open tube. Suppose someone has taken a long hoselike tube, coiled it up in a box, and let one open end stick out one side of the box and the other out the other. You are not allowed to see how much of the tube is coiled inside the box. By adding a small tuning trombone to a protruding end, you find that you get a resonance at \(523.3\) cps from your tuning fork. That means that the total length is either \(\frac{1}{2} \lambda\), or \(\lambda\), or \(\frac{A}{2} \lambda\), or \(\ldots . .\) How can you find out whether the tube is an odd or even number of half-wavelengths? Hold two vibrating forks at one end of the tube and listen to the beats. Get the rhythm in your head so that if you remove one fork momentarily and then replace it (without disturbing the continued vibrations of both forks), you can tell that the beat maximum comes "on the beat" (in musical jargon) just where it should be. Practice several times so that you can skip a beat, count beats in your head, and come back in step when you replace the fork. (You can adjust the rubber-band loading to get a convenient beat frequency. If you find all this difficult, you can use a metronome.) Now! This time, instead of replacing the (momentarily) removed fork at the same end of the tube, carry it to the other end. Again listen for the beats. (Both forks have continued vibrating all this time.) Do they come back "on the beat," or do they come back "on the off-beat"? Depending on the experimental result, you should be able to decide whether the tube is an odd or even number of half-wavelengths. Predict the answer; then try the experiment with your half-wavelength tube. (Make another tube one wavelength long to get the opposite result.)

Transitory standing waves on a slinky. Attach one end of a stinky to a telephone pole or something. Hold the other end. Stretch the slinky out to \(30 \mathrm{ft}\) or so. Shake the end of the slinky about 3 or 4 times as rapidly as you can. \(\mathrm{A}^{\text {" } \text { wave packet" }}\) " is thus propagated down the slinky. After you have sufficiently enjoyed following packets back and forth, try something new: This time, keep your attention fixed on a region near the fixed end of the slinky. As the packet comes in, reflects, and returns, you should see transitory standing waves during the time interval in which the incident and reflected wave packets overlap. (It may help to fix both ends of the slinky so that you can watch the process at close range at your end of the slinky.) That should help to convince you that a standing wave can always be regarded as the superposition of two traveling waves traveling in opposite directions.

Impedance matching by "tapered" index of refraction. Suppose you want to match optical impedances between a region of index \(n_{1}\) and a region of index \(n_{2}\), and you want to expend a total distance \(L\) in the impedance-matching transition region. What is the optimum \(z\) dependence of the index \(n\) between the two regions? Is it exponential? Why not?

Overtones in tuning fork. Does your C523.3 tuning fork emit nothing but a 523 -cps sound? Strike the fork against something hard. You should hear a faint high tone in addition to the strong 523 -cps tone. The high tone dies away in two or three seconds. It is a higher mode of the fork and is strongly damped because it involves greater bending of the prongs. What about the note an octave higher, Cl048? This is difficult to listen for because of the presence of the fundamental, C523. To search for it, use a resonating tube. Tune a tube to \(\mathrm{C} 1046\) by tapping it on your head and listening for the octave above C523. (Or simply cut it by "theory," subtracting \(0.6\) of a radius \(R\) for each end from \(\lambda / 2\) to get the length.) Hold the \(\mathrm{C} 523\) fork at the end of the C1046 tube and listen. (Use a tube tuned at C523 as a control. Move the fork back and forth between the C523 and C1046 tubes.)

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