You and a friend are holding the two ends of a Slinky stretched out between you. How would you move your end of the Slinky to create (a) transverse waves or (b) longitudinal waves?

Short Answer

Expert verified
Answer: To create a transverse wave with a Slinky, move your end up and down perpendicular to the direction of the stretched Slinky, causing a horizontal wave to travel through it. For a longitudinal wave, move your end back and forth parallel to the stretched Slinky, causing it to compress and expand as the wave moves through it.

Step by step solution

01

Creating Transverse Waves

To create transverse waves in a Slinky, you will need to move your end of the Slinky up and down (perpendicular to the direction of the stretched Slinky). This motion will cause parts of the Slinky to move up and down as well, creating a wave that moves through the Slinky horizontally. The key motion here is to move your hand in an up and down motion without moving it horizontally.
02

Creating Longitudinal Waves

To create longitudinal waves in a Slinky, you will need to move your end of the Slinky back and forth (parallel to the direction of the stretched Slinky). This motion will cause parts of the Slinky to compress and expand as the wave moves through it. The key motion here is to move your hand in a back and forth motion along the same line as the stretched Slinky, without moving it up and down.

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

Students in a lab produce standing waves on stretched strings connected to vibration generators. One such wave is described by the wave function \(y(x, t)=(2.00 \mathrm{~cm}) \sin \left[\left(20.0 \mathrm{~m}^{-1}\right) x\right] \cos \left[\left(150 . \mathrm{s}^{-1}\right) t\right],\) where \(y\) is the transverse displacement of the string, \(x\) is the position along the string, and \(t\) is time. Rewrite this wave function in the form for a right- moving and a left-moving wave: \(y(x, t)=\) \(f(x-v t)+g(x+v t)\); that is, find the functions \(f\) and \(g\) and the speed, \(v\)

A particular steel guitar string has mass per unit length of \(1.93 \mathrm{~g} / \mathrm{m}\). a) If the tension on this string is \(62.2 \mathrm{~N},\) what is the wave speed on the string? b) For the wave speed to be increased by \(1.0 \%\), how much should the tension be changed?

If two traveling waves have the same wavelength, frequency, and amplitude and are added appropriately, the result is a standing wave. Is it possible to combine two standing waves in some way to give a traveling wave?

Consider a linear array of \(n\) masses, each equal to \(m,\) connected by \(n+1\) springs, all massless and having spring constant \(k\), with the outer ends of the first and last springs fixed. The masses can move without friction in the linear dimension of the array. a) Write the equations of motion for the masses. b) Configurations of motion for which all parts of a system oscillate with the same angular frequency are called normal modes of the system; the corresponding angular frequencies are the system's normal-mode angular frequencies. Find the normal-mode angular frequencies of this array.

What is the wave speed along a brass wire with a radius of \(0.500 \mathrm{~mm}\) stretched at a tension of \(125 \mathrm{~N}\) ? The density of brass is \(8.60 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).

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