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A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.120 \(\mathrm{m} .\) The maximum speed of the block is 3.90 \(\mathrm{m} / \mathrm{s}\) . What is the maximum magnitude of the acceleration of the block?

Short Answer

Expert verified
The maximum acceleration is approximately 127 m/s².

Step by step solution

01

Identify Given Values

We are given the amplitude \( A = 0.120 \, \text{m} \) and the maximum speed \( v_{max} = 3.90 \, \text{m/s} \). We need to find the maximum acceleration \( a_{max} \).
02

Recall SHM Relationship

In simple harmonic motion, the maximum speed \( v_{max} \) is related to the angular frequency \( \omega \) and amplitude \( A \) by the formula: \[ v_{max} = A \omega \]
03

Solve for Angular Frequency \( \omega \)

Rearrange the formula from Step 2 to solve for \( \omega \): \[ \omega = \frac{v_{max}}{A} = \frac{3.90 \, \text{m/s}}{0.120 \, \text{m}} = 32.5 \, \text{rad/s} \]
04

Use Maximum Acceleration Formula

The maximum acceleration \( a_{max} \) in SHM is given by \[ a_{max} = A \omega^2 \]
05

Calculate \( a_{max} \)

Substitute for \( A \) and \( \omega \) into the acceleration formula: \[ a_{max} = 0.120 \, \text{m} \times (32.5 \, \text{rad/s})^2 = 126.75 \, \text{m/s}^2 \]
06

Round the Answer

Round the acceleration to a reasonable number of significant figures or as needed: \[ a_{max} \approx 127 \, \text{m/s}^2 \]

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Key Concepts

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

Maximum Speed in SHM
In simple harmonic motion (SHM), an object oscillates back and forth in a regular, repeating manner. The maximum speed, or peak velocity, of an object in SHM can be significant when determining other properties of its motion. The maximum speed in SHM occurs as the object passes through the equilibrium position. At this instant, the entire energy of the system is kinetic, which explains why it reaches its highest value.

The relationship between maximum speed, amplitude, and angular frequency is crucial here. We express it mathematically as:
  • \(v_{max} = A \omega\)
where:
  • \(v_{max}\) is the maximum speed.
  • \(A\) stands for amplitude, the maximum displacement from the equilibrium position.
  • \(\omega\) denotes the angular frequency, which describes how fast the oscillation occurs.
Understanding this relationship helps in determining other parameters of SHM, such as angular frequency and maximum acceleration.
Angular Frequency
Angular frequency \(\omega\) in simple harmonic motion represents how quickly the object oscillates. It is a measure of rotation per unit time, even when dealing with linear oscillations like those observed in springs or pendulums.

To find the angular frequency of a system, we rearrange the formula linking maximum speed, angular frequency, and amplitude:
  • \(\omega = \frac{v_{max}}{A}\)
This equation allows us to calculate the angular frequency directly when the maximum speed and amplitude are known.

In the provided exercise,
  • \(v_{max} = 3.90 \, \text{m/s}\)
  • \(A = 0.120 \, \text{m}\)
By substituting these values, we calculated \(\omega\) as \(32.5 \, \text{rad/s}\). Understanding \(\omega\) not only describes the tempo of the periodic motion but also links directly to determining other motion characteristics like maximum acceleration.
Maximum Acceleration in SHM
Maximum acceleration in simple harmonic motion is the highest rate of change of velocity as the system oscillates. This typically occurs when the object is at its maximum displacement from the equilibrium position. Calculating maximum acceleration is key to understanding the dynamics and forces involved in SHM, which can describe phenomena like spring systems and pendulums.

The formula for maximum acceleration is:
  • \(a_{max} = A \omega^2\)
In this formula:
  • \(a_{max}\) represents the maximum acceleration.
  • \(A\) is the amplitude.
  • \(\omega\) is the angular frequency.
Following the provided exercise example, substituting \(A = 0.120 \, \text{m}\) and \(\omega = 32.5 \, \text{rad/s}\) into the equation results in \(a_{max} = 126.75 \, \text{m/s}^2\). For practical purposes, this can be rounded to approximately \(127 \, \text{m/s}^2\).

This calculation highlights how amplitude and angular frequency directly influence acceleration, thus helping to predict the object's behavior in SHM systems.

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Most popular questions from this chapter

CP A rifle bullet with mass 8.00 \(\mathrm{g}\) and initial horizontal velocity 280 \(\mathrm{m} / \mathrm{s}\) strikes and embeds itself in a block with mass 0.992 \(\mathrm{kg}\) that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0 \(\mathrm{cm}\) . After the impact, the block moves in SHM. Calculate the period of this motion.

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom- pom's coordinate is \(x=+9.0 \mathrm{cm} ;(\mathrm{c})\) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach used in Section \(14.3,\) and which cannot? Explain.

Inside a NASA test vehicle, a \(3.50-\) kg ball is pulled along by a horizontal ideal spring fixed to a friction-free table. The force constant of the spring is 225 \(\mathrm{N} / \mathrm{m}\) . The vehicle has a steady acceleration of \(5.00 \mathrm{m} / \mathrm{s}^{2},\) and the ball is not oscillating. Suddenly, when the vehicle's speed has reached \(45.0 \mathrm{m} / \mathrm{s},\) its engines turn off, thus eliminating its acceleration but not its velocity. Find (a) the amplitude and (b) the frequency of the resulting oscillations of the ball. (c) What will be the ball's maximum speed relative to the vehicle?

CALC Jerk. A guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

In a physics lab, you attach a \(0.200-\mathrm{kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

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