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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/s2. 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=ω2x, where a is the acceleration, ω is the angular frequency, and x is the displacement from equilibrium.
03

Solve for Angular Frequency

We have a=5.30 m/s2 and x=0.280 m. Using the formula a=ω2x, we can solve for ω as follows:ω2=ax=5.300.28018.93 s2Thus, ω=18.934.35 rad/s.
04

Convert Angular Frequency to Frequency

The relationship between angular frequency ω and frequency f is given by ω=2πf. Solving for f, we have:f=ω2π=4.352π0.692 Hz

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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 ω=2πf, where ω is the angular frequency. To find the frequency from the angular frequency, you simply rearrange the formula:
  • Substitute the value of angular frequency (ω) into f=ω2π
  • 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** ω 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=ω2x. Here, a is the acceleration, x is the displacement from the equilibrium position, and ω is the angular frequency. From this, you can solve for angular frequency as follows:
  • Rearrange the formula to ω2=ax
  • Compute ω by taking the square root of ax.
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=ω2x 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 ω. Lastly, if the problem requires, convert these findings using other relationships, such as ω=2πf to find frequency.

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

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Most popular questions from this chapter

You pull a simple pendulum 0.240 m long to the side through an angle of 3.50 and release it. (a) How much time does it take the pendulum bob to reach its highest speed? (b) How much time does it take if the pendulum is released at an angle of 1.75 instead of 3.50?

A 40.0-N force stretches a vertical spring 0.250 m. (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 s? (b) If the amplitude of the motion is 0.050 m and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

A 2.00-kg bucket containing 10.0 kg of water is hanging from a vertical ideal spring of force constant 450 N/m and oscillating up and down with an amplitude of 3.00 cm. Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 g/s. When the bucket is half full, find (a) the period of oscillation and (b) the rate at which the period is changing with respect to time. Is the period getting longer or shorter? (c) What is the shortest period this system can have?

A 1.50-kg mass on a spring has displacement as a function of time given by x(t)=7.40cm)cos[(4.16rad/s)t2.42] Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at t= 1.00 s; (f) the force on the mass at that time.

A 0.500-kg glider, attached to the end of an ideal spring with force constant k= 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at x=0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at x=0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

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