A (somewhat risky) way of telling if a train that cannot be seen or heard is approaching is by placing your ear on the rail. Explain why this works.

Short Answer

Expert verified
Answer: Placing your ear on a railway track helps detect an approaching train because sound waves travel faster and more efficiently through the solid steel track than through air, allowing you to sense the vibrations created by the train's movement even if it's too far away to be heard through the air. However, this method is not recommended for safety reasons as it puts individuals at risk of being too close to the tracks when a train approaches.

Step by step solution

01

Understanding sound waves

Sound is a vibrational energy transmitted through mediums like air, water, and solids. The vibration causes the particles in the medium to oscillate back and forth, creating sound waves. These sound waves are longitudinal waves that travel at different speeds depending on the medium.
02

Properties of solids for sound transmission

In solids, particles are more densely packed than in air or water, allowing them to transmit sound waves faster and more efficiently. The speed of sound in steel, which is the primary material used in railway tracks, is around 5,120 meters per second (m/s), while in air, it is approximately 343 m/s. Thus, sound travels significantly faster through steel than through air.
03

Detecting vibrations through the ear

Our ears can detect the vibrations created by sound waves in the environment. When you place your ear on a solid surface, like a railway track, you can effectively detect the vibrations that travel along the track. This means that if a train is approaching, even from a distance, the vibrations from the train's movement will propagate through the steel track and your ear will sense the vibrations more easily when in direct contact with the track.
04

Why it works for detecting approaching trains

Placing your ear on the railway track allows you to effectively detect an approaching train because the solid track can transmit the sound waves faster and more efficiently than through air. As a result, you can sense the vibrations created by the train's movement even if it's too far away to be heard through the air. This helps in determining if a train is coming without actually seeing or hearing it through the air, although it is a risky method and not recommended for safety reasons.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two 100.0-W speakers, A and B, are separated by a distance \(D=3.6 \mathrm{~m} .\) The speakers emit in-phase sound waves at a frequency \(f=10,000.0 \mathrm{~Hz}\). Point \(P_{1}\) is located at \(x_{1}=4.50 \mathrm{~m}\) and \(y_{1}=0 \mathrm{~m} ;\) point \(P_{2}\) is located at \(x_{2}=4.50 \mathrm{~m}\) and \(y_{2}=-\Delta y .\) Neglecting speaker \(\mathrm{B}\), what is the intensity, \(I_{\mathrm{A} 1}\) (in \(\mathrm{W} / \mathrm{m}^{2}\) ), of the sound at point \(P_{1}\) due to speaker \(\mathrm{A}\) ? Assume that the sound from the speaker is emitted uniformly in all directions. What is this intensity in terms of decibels (sound level, \(\beta_{\mathrm{A} 1}\) )? When both speakers are turned on, there is a maximum in their combined intensities at \(P_{1} .\) As one moves toward \(P_{2},\) this intensity reaches a single minimum and then becomes maximized again at \(P_{2}\). How far is \(P_{2}\) from \(P_{1},\) that is, what is \(\Delta y ?\) You may assume that \(L \gg \Delta y\) and that \(D \gg \Delta y\), which will allow you to simplify the algebra by using \(\sqrt{a \pm b} \approx a^{1 / 2} \pm \frac{b}{2 a^{1 / 2}}\) when \(a \gg b\).

You are creating a sound wave by shaking a paddle in a liquid medium. How can you increase the speed of the resulting sound wave? a) You shake the paddle harder to give the medium more kinetic energy. b) You vibrate the paddle more rapidly to increase the frequency of the wave. c) You create a resonance with a faster-moving wave in air. d) All of these will work. e) None of these will work. f) Only (a) and (b) will work. g) Only (a) and (c) will work. h) Only (b) and (c) will work.

On a windy day, a child standing outside a school hears the school bell. If the wind is blowing toward the child from the direction of the bell, will it alter the frequency, the wavelength, or the velocity of the sound heard by the child?

You are traveling in a car toward a hill at a speed of \(40.0 \mathrm{mph} .\) The car's horn emits sound waves of frequency \(250 \mathrm{~Hz},\) which move with a speed of \(340 \mathrm{~m} / \mathrm{s}\) a) Determine the frequency with which the waves strike the hill. b) What is the frequency of the reflected sound waves you hear? c) What is the beat frequency produced by the direct and the reflected sounds at your ears?

The Moon has no atmosphere. Is it possible to generate sound waves on the Moon?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free