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Chapter 11: Problem 33

A 2.5-kg body undergoes SHM and makes exactly 3 vibrations each second. Compute the acceleration and the restoring force acting on the body when its displacement from the equilibrium position is \(5.0 \mathrm{~cm}\).

Short Answer

Expert verified
Acceleration is approximately 17.76 m/s虏, and restoring force is 44.4 N.

Step by step solution

01

Define the Known Variables

First, let's identify the information given in the problem:- Mass of the body, \( m = 2.5 \) kg.- Frequency of vibration, \( f = 3 \) Hz.- Displacement from equilibrium position, \( x = 5.0 \) cm (or 0.05 m).
02

Calculate the Angular Frequency

The angular frequency \( \omega \) is related to frequency by the formula \( \omega = 2 \pi f \). Substituting the known value of \( f \):\[ \omega = 2 \pi \times 3 = 6 \pi \text{ rad/s} \].
03

Calculate the Maximum Acceleration

The maximum acceleration \( a_{max} \) in simple harmonic motion is given by \( a_{max} = \omega^2 x \). Using the values \( \omega = 6\pi \) and \( x = 0.05 \) m:\[ a_{max} = (6\pi)^2 \times 0.05 = 36\pi^2 \times 0.05 \approx 17.76 \text{ m/s}^2 \].
04

Calculate the Restoring Force

The restoring force \( F \) is given by Hooke's Law for SHM: \( F = -kx \), where \( k \) is the spring constant, and can also be found by \( a = \omega^2 x \) leading to \( F = ma = m \omega^2 x \). Substituting the values, \( m = 2.5 \), \( \omega^2 = 36\pi^2 \), \( x = 0.05 \):\[ F = 2.5 \times 17.76 \approx 44.4 \text{ N} \].

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Key Concepts

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

Understanding Restoring Force in SHM
The idea of a restoring force is fundamental in understanding simple harmonic motion (SHM). This force is the key that drives an object back to its equilibrium position. It acts in the opposite direction of displacement. In simple harmonic motion, the restoring force is directly proportional to the displacement from equilibrium but acts in the opposite direction. This relationship is mathematically expressed as:
  • Restoring force, \( F = -kx \). Here, \( k \) is a constant that varies depending on the system and is known as the spring constant.

The negative sign indicates that the force is in the opposite direction to the displacement.
In our exercise, the body's restoring force is calculated using Hooke's Law combined with the body's mass and angular frequency to find the spring constant indirectly. This becomes:
  • Restoring force, \( F = m \omega^2 x \).

Plug in the available values, and you get a restoring force of approximately 44.4 N, acting towards the equilibrium position.
Calculating Angular Frequency
In simple harmonic motion, the concept of angular frequency \( \omega \) describes how quickly an object oscillates around its equilibrium point. Angular frequency is measured in radians per second, a crucial measure of SHM.
It links directly to the ordinary frequency \( f \) (measured in hertz) through the formula:
  • \( \omega = 2 \pi f \).

This formula is derived from the fact that one complete cycle in radians is \( 2\pi \) times the basic frequency.
In our example, since the frequency is given as 3 vibrations per second:
  • Angular frequency, \( \omega = 2 \pi \times 3 = 6 \pi \) rad/s.

This angular frequency then feeds into other calculations, helping to define how rapidly the system can oscillate.
Determining Maximum Acceleration
In simple harmonic motion, the maximum acceleration provides insight into the dynamic behavior of the oscillating system. When displacement from equilibrium is at its peak, so is the force acting to pull it back, which means higher acceleration.
The formula marking this is:
  • Maximum acceleration, \( a_{max} = \omega^2 x \).

Here, \( \omega^2 \) shows the relation between angular frequency and the speed of returning the object to its resting state.
For the provided scenario, using the angular frequency \( \omega = 6\pi \) rad/s and displacement \( x = 0.05 \) meters, the calculation yields:
  • \( a_{max} = (6\pi)^2 \times 0.05 \approx 17.76 \text{ m/s}^2 \).

This acceleration at maximum displacement illustrates the swift pace at which the system tries to restore balance.

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Most popular questions from this chapter

A pendulum is timed as it swings back and forth. The clock is started when the bob is at the left end of its swing. When the bob returns to the left end for the 90 th return, the clock reads \(60.0 \mathrm{~s}\). What is the period of vibration? The frequency?

Compute the acceleration due to gravity at a place where a simple pendulum \(150.3 \mathrm{~cm}\) long swings through \(100.0\) cycles in \(246.7\) \(\mathrm{S} .\) We have $$ T=\frac{246.7 \mathrm{~s}}{100.0}=2.467 \mathrm{~s} $$ Squaring \(T=2 \pi \sqrt{L / g}\) and solving for \(g\) yields $$ g=\frac{4 \pi^{2}}{T^{2}} L=9.749 \mathrm{~m} / \mathrm{s}^{2} $$

A 200-g mass vibrates horizontally without friction at the end of a horizontal spring for which \(k=7.0 \mathrm{~N} / \mathrm{m} .\) The mass is displaced \(5.0\) \(\mathrm{cm}\) from equilibrium and released. Find \((a)\) its maximum speed and \((b)\) its speed when it is \(3.0 \mathrm{~cm}\) from equilibrium. ( \(c\) ) What is its acceleration in each of these cases? where \(k=7.0 \mathrm{~N} / \mathrm{m}, x_{0}=0.050 \mathrm{~m}\), and \(m=0.200 \mathrm{~kg} .\) Solving for \(|v|\) gives $$ |v|=\sqrt{\left(x_{0}^{2}-x^{2}\right) \frac{k}{m}} $$ (a) The speed is a maximum when \(x=0\); that is, when the mass is passing through the equilibrium position: $$ |v|=x_{0} \sqrt{\frac{k}{m}}=(0.050 \mathrm{~m}) \sqrt{\frac{7.0 \mathrm{~N} / \mathrm{m}}{0.200 \mathrm{~kg}}}=0.30 \mathrm{~m} / \mathrm{s} $$ (b) When \(x=0.030 \mathrm{~m}\), $$ |v|=\sqrt{\frac{7.0 \mathrm{~N} / \mathrm{m}}{0.200 \mathrm{~kg}}\left[(0.050)^{2}-(0.030)^{2}\right] \mathrm{m}^{2}}=0.24 \mathrm{~m} / \mathrm{s} $$ (c) Using \(F=m a\) and \(F=k x\), $$ a=\frac{k}{m} x=\left(35 \mathrm{~s}^{-2}\right)(x) $$ which yields \(a=0\) when the mass is at \(x=0\), and \(a=1.1 \mathrm{~ms}^{2}\) when \(x=0.030 \mathrm{~m}\).

How much energy is stored in a spring with an elastic constant of \(1000 \mathrm{~N} / \mathrm{m}\) when it is compressed \(10 \mathrm{~cm}\) ?

A 300 -g object attached to the end of a spring oscillates with an amplitude of \(7.0 \mathrm{~cm}\) and a frequency of \(1.80 \mathrm{~Hz}\). (a) Find its maximum speed and maximum acceleration. (b) What is its speed when it is \(3.0 \mathrm{~cm}\) from its equilibrium position?
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