Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the block is at \(x =\) 0.280 m, the acceleration of the block is \(-\)5.30 m/s\(^2\). What is the frequency of the motion?

Short Answer

Expert verified
The frequency of the motion is approximately 0.692 Hz.

Step by step solution

01

Understand the Problem

We need to find the frequency of the block attached to a spring moving in simple harmonic motion (SHM) given the position and acceleration.
02

Recall the Formula for Acceleration in SHM

In SHM, the acceleration of the block is given by the formula \( a = -\omega^2 x \), where \( a \) is the acceleration, \( \omega \) is the angular frequency, and \( x \) is the displacement from equilibrium.
03

Solve for Angular Frequency

We have \( a = -5.30 \text{ m/s}^2 \) and \( x = 0.280 \text{ m} \). Using the formula \( a = -\omega^2 x \), we can solve for \( \omega \) as follows:\[\omega^2 = \frac{a}{x} = \frac{-5.30}{-0.280} \approx 18.93 \text{ s}^{-2}\]Thus, \( \omega = \sqrt{18.93} \approx 4.35 \text{ rad/s}\).
04

Convert Angular Frequency to Frequency

The relationship between angular frequency \( \omega \) and frequency \( f \) is given by \( \omega = 2 \pi f \). Solving for \( f \), we have:\[f = \frac{\omega}{2 \pi} = \frac{4.35}{2 \pi} \approx 0.692 \text{ Hz}\]

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.

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

Key Concepts

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

Frequency Calculation
Simple harmonic motion (SHM) is a fascinating topic in physics, describing oscillatory motions such as the movement of a block attached to a spring. One of the key characteristics we often want to determine in SHM is the **frequency** of the motion. Frequency, denoted by \(f\), refers to how often the motion completes a full cycle each second and is measured in Hertz (Hz).

In the context of a spring-mass system undergoing SHM, the frequency can be deduced using the relationship between angular frequency and ordinary frequency. The **relationship** is given by \(\omega = 2\pi f\), where \(\omega\) is the angular frequency. To find the frequency from the angular frequency, you simply rearrange the formula:
  • Substitute the value of angular frequency (\(\omega\)) into \(f = \frac{\omega}{2\pi}\)
  • Perform the division to get the frequency in Hz.
By understanding and applying this straightforward method, you can calculate the frequency of any SHM system once you know its angular frequency.
Angular Frequency
**Angular frequency** \(\omega\) plays a crucial role in the analysis of simple harmonic motion. It can be thought of as a measure of how fast an object swings through its cycle. Unlike regular frequency, angular frequency is expressed in radians per second (rad/s).

To find the angular frequency in a spring-mass system, you can use the formula that connects acceleration, angular frequency, and displacement: \(a = -\omega^2 x\). Here, \(a\) is the acceleration, \(x\) is the displacement from the equilibrium position, and \(\omega\) is the angular frequency. From this, you can solve for angular frequency as follows:
  • Rearrange the formula to \(\omega^2 = \frac{a}{x}\)
  • Compute \(\omega\) by taking the square root of \(\frac{a}{x}\).
This calculation helps to find the natural tendency of the system's oscillations, providing essential insight into the motion's dynamics.
Physics Problem Solving
When dealing with physics problems like simple harmonic motion, a systematic approach to **problem-solving** can simplify finding the solution. Here is a recommended method:

First, **understand** the problem by identifying known quantities and what needs to be found. In problems involving SHM, often you know the displacement and the acceleration.

Next, **recall relevant formulas**. For SHM, a valuable formula is \(a = -\omega^2 x\) which connects the acceleration to angular frequency and displacement.

Then, **substitute** known values into the formula to solve for the unknowns. If you're solving for angular frequency, rearrange and compute as necessary to find \(\omega\). Lastly, if the problem requires, convert these findings using other relationships, such as \(\omega = 2\pi f\) to find frequency.

This step-by-step process combines logical reasoning with mathematical manipulation, enabling you to tackle various physics problems effectively.

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

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

A 2.00-kg frictionless block is attached to an ideal spring with force constant 315 N/m. Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude of the motion, (b) the block's maximum acceleration, and (c) the maximum force the spring exerts on the block.

A small sphere with mass \(m\) is attached to a massless rod of length \(L\) that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle \(\theta\) from the vertical, and released from rest. (a) In a diagram, show the pendulum just after it is released. Draw vectors representing the \(forces\) acting on the small sphere and the \(acceleration\) of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere? (b) Repeat part (a) for the instant when the pendulum rod is at an angle \(\theta\)/2 from the vertical. (c) Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

An object is undergoing SHM with period 1.200 s and amplitude 0.600 m. At \(t =\) 0 the object is at \(x =\) 0 and is moving in the negative \(x\)-direction. How far is the object from the equilibrium position when \(t =\) 0.480 s?

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 m/s to the right and an acceleration of 8.40 m/s\(^2\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

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