Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 47

The displacement of an object in SHM is given by $y(t)=(8.0 \mathrm{cm}) \sin [(1.57 \mathrm{rad} / \mathrm{s}) t] .$ What is the frequency of the oscillations?

Short Answer

Expert verified
Answer: The frequency of the oscillations is approximately \(0.25\,\mathrm{Hz}\).

Step by step solution

01

Identify the equation of motion

The displacement of the object in SHM is given by \(y(t) = (8.0 \mathrm{cm})\sin[(1.57 \mathrm{rad} / \mathrm{s})t]\). This equation has the general form \(y(t) = A\sin(\omega t)\), where A is the amplitude and ω is the angular frequency.
02

Extract the angular frequency

From the given equation, we can see that the angular frequency ω is equal to \(1.57 \mathrm{rad} / \mathrm{s}\).
03

Find the frequency

We know the relation between angular frequency (ω) and frequency (f) is given as \(\omega = 2 \pi f\). We need to solve this equation for f. $$ f = \frac{\omega}{2 \pi} $$ Now, substitute ω with the value we found earlier: $$ f = \frac{1.57 \mathrm{rad}/\mathrm{s}}{2 \pi} $$ Calculate the frequency f: $$ f \approx 0.25\,\mathrm{Hz} $$ The frequency of the oscillations is approximately \(0.25\,\mathrm{Hz}\).

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.

Start your free trial

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

The period of oscillation of a spring-and-mass system is \(0.50 \mathrm{s}\) and the amplitude is \(5.0 \mathrm{cm} .\) What is the magnitude of the acceleration at the point of maximum extension of the spring?
An empty cart, tied between two ideal springs, oscillates with $\omega=10.0 \mathrm{rad} / \mathrm{s} .$ A load is placed in the cart. making the total mass 4.0 times what it was before. What is the new value of \(\omega ?\)
A cart with mass \(m\) is attached between two ideal springs, each with the same spring constant \(k\). Assume that the cart can oscillate without friction. (a) When the cart is displaced by a small distance \(x\) from its equilibrium position, what force magnitude acts on the cart? (b) What is the angular frequency, in terms of \(m, x,\) and \(k,\) for this cart? (W) tutorial: cart between springs) (IMAGE NOT COPY)
A \(0.50-\mathrm{kg}\) object, suspended from an ideal spring of spring constant \(25 \mathrm{N} / \mathrm{m},\) is oscillating vertically. How much change of kinetic energy occurs while the object moves from the equilibrium position to a point \(5.0 \mathrm{cm}\) lower?
The ratio of the tensile (or compressive) strength to the density of a material is a measure of how strong the material is "pound for pound." (a) Compare tendon (tensile strength \(80.0 \mathrm{MPa}\), density $1100 \mathrm{kg} / \mathrm{m}^{3}\( ) with steel (tensile strength \)\left.0.50 \mathrm{GPa}, \text { density } 7700 \mathrm{kg} / \mathrm{m}^{3}\right)$ which is stronger "pound for pound" under tension? (b) Compare bone (compressive strength \(160 \mathrm{MPa}\), density $1600 \mathrm{kg} / \mathrm{m}^{3}$ ) with concrete (compressive strength $\left.0.40 \mathrm{GPa}, \text { density } 2700 \mathrm{kg} / \mathrm{m}^{3}\right):$ which is stronger "pound for pound" under compression?
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Turning Points in Physics

Read Explanation

Classical Mechanics

Read Explanation

Medical Physics

Read Explanation

Space Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
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