Question

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

The frequency of the motion is approximately 0.692 Hz.

Step-by-step solution

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.

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.

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, $\omega =\sqrt{18.93}\approx 4.35\text{ rad/s}$.

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:$$f=\frac{\omega }{2\pi }=\frac{4.35}{2\pi }\approx 0.692\text{ Hz}$$

Key concepts

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.