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A \(342-\mathrm{g}\) mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is \(18.6 \mathrm{~m} / \mathrm{s}^{2}\), and its maximum speed is \(1.75 \mathrm{~m} / \mathrm{s}\). Determine (a) the angular frequency, (b) the amplitude, and (c) the spring constant.

Short Answer

Expert verified
The angular frequency is \(\omega = 0.094 rad/s\), \nthe amplitude is \(A = 18.6 m\), \nand the spring constant is \(k = 0.92 N/m\).

Step by step solution

01

Calculate Angular Frequency

Angular frequency \(\omega\) is given by the formula \(\omega = \frac{v_{max}}{a_{max}}\), where \(v_{max}\) is the maximum speed and \(a_{max}\) is the maximum acceleration. Substitute the given values into the formula to find the angular frequency.
02

Determine the Amplitude

The amplitude A can be found by using the formula \(A = \frac{v_{max}}{\omega}\), where \(v_{max}\) is the maximum speed and \(\omega\) is the angular frequency. Substitute the values obtained from Step 1 into the formula to find the the amplitude.
03

Find the Spring Constant

The spring constant \(k\) can be calculated by using the formula \(k = m\omega^2\), where m is the mass and \(\omega\) is the angular frequency. Substitute the values obtained from Steps 1 and the given value for mass into the formula to find the spring constant.

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Key Concepts

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

Angular Frequency
Angular frequency is a key parameter in understanding the behavior of simple harmonic motion. It is represented by the Greek letter omega \(\omega\) and it tells us how many oscillations an object makes per unit of time. The angular frequency is related to other parameters such as maximum speed \(v_{max}\) and maximum acceleration \(a_{max}\), reflecting how quickly the mass moves back and forth on the spring.

For a body undergoing simple harmonic motion, the maximum speed and acceleration are useful in determining the angular frequency. We assume that the motion obeys Hooke's Law, which states that the force on the mass is linearly dependent on the displacement from the equilibrium position, and is always directed towards that equilibrium point. This results in angular frequency being a measure not just of time but also of the rate of change of displacement with respect to time.
Amplitude Calculation
Amplitude is a term that refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In the context of simple harmonic motion, the amplitude \(A\) signifies the furthest distance from the equilibrium position that the mass reaches during its motion along the spring.

Amplitude is crucial because it represents the maximum displacement from the center position and can also define the energy in the system. The calculation of amplitude uses angular frequency and the maximum speed of the oscillating object. With gravity being a constant force and assuming there is no energy loss, the amplitude remains constant as well, which is a defining characteristic of simple harmonic motion.
Spring Constant
The spring constant, denoted by \(k\), is a value that quantifies the stiffness of a spring. It's key to understanding how much force is needed to compress or extend a spring by a certain distance. In the case of simple harmonic motion, the spring constant plays an essential role as it determines the restoring force which is pushing the mass back towards the equilibrium point.

The spring constant is directly related to the angular frequency and the mass of the object attached to the spring. A higher spring constant means a stiffer spring, leading to a higher frequency of oscillation for a given mass. It's indirectly connected to energy as well since the potential energy stored in a spring during compression or elongation is proportional to the spring constant.
Maximum Acceleration
Maximum acceleration in the context of simple harmonic motion is the greatest acceleration value achieved by the oscillating object, occurring when the object passes through the equilibrium position. The formula for maximum acceleration is \(a_{max} = \omega^2 \times A\), where \(\omega\) is the angular frequency and \(A\) is the amplitude of the oscillation.

This acceleration is the result of the restoring force exerted by the spring on the mass. It's significant because it dictates how quickly the object's velocity changes as it moves. Notably, the greater the maximum acceleration, the more powerful the force that returns the mass to its equilibrium position, which, in turn, causes a faster oscillation.
Maximum Speed
In simple harmonic motion, maximum speed \(v_{max}\) is the greatest speed the oscillating object reaches, which happens as it crosses the equilibrium point. This is because at the equilibrium position, all of the energy in the system is kinetic. The value of maximum speed can be calculated with the equation \(v_{max} = \omega \times A\), which directly ties it to both the amplitude and the angular frequency.

The maximum speed is an essential aspect of oscillatory motion as it tells us how fast the object is moving at its quickest point. When you know the maximum speed, you can gain insights about the kinetic energy of the system at different points during the oscillation. Moreover, the maximum speed contributes to understanding the total energy conserved in the harmonic motion, which is the sum of both potential energy in the spring and kinetic energy of the mass.

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Most popular questions from this chapter

This problem explores the nonlinear pendulum discussed P qualitatively in Conceptual Example 13.1. You can tackle this problem if you have experience with your calculator's differential-equation solving capabilities or if you've used a software program like Mathematica or Maple that can solve differential equations numerically. On page 252 we wrote Newton's law for a pendulum as \(1 d^{2} \theta / d t^{2}=-m g L \sin \theta\). (a) Rewrite this equation in a form suitable for a simple pendulum, but without making the approximation \(\sin \theta \cong \theta\). Although it won't affect the form of the equation, assume that your pendulum uses a massless rigid rod rather than a string, so it can turn completely upside down without collapsing. (b) Enter your equation into your calculator or software, and produce graphical solutions to the equation for the situation where you specify the initial kinetic energy \(K_{0}\) when the pendulum is at its bottommost position. In particular, describe solutions for (i) \(K_{0}\) \(K_{0} \leqslant U_{\max }\), and (iii) \(K_{0}>U_{\max } .\) Here \(U_{\max }\) is the maximum pos- (ii) \(K_{0} \leqslant U_{\max }\), and (iii) \(K_{0}>U_{\max } .\) Here \(U_{\max }\) is the maximum pos- sible potential energy for the system, which occurs when the pendulum is completely upside down; \(U_{0}=2 L m g\), where \(L\) is the pendulum's length.

This problem explores what would happen if a hole were drilled through Earth's center and out the other side, and an object were dropped into the hole. Approximating Earth as a uniform solid sphere, the gravitational acceleration within the planet (including inside the hypothetical hole) would be \(g(r)=g_{0}\left(r / R_{\mathrm{E}}\right)\), where \(g_{0}\) is the value at Earth's surface, \(r\) is the distance from Earth's center, and \(R_{\mathrm{E}}\) is Earth's radius. This gravitational acceleration is directed toward Earth's center. (a) Write an expression for the force on a mass \(m\) at any point \(r\) in the hole, apply Newton's second law, and show that you get an equation analogous to Equation 13.3. Neglect air resistance. (b) Use your analogy to find an expression for the period of the simple harmonic motion that results when the mass is dropped into the hole. (c) Use appropriate values to find a numerical value for the period, and compare with the period for circular low- Earth orbit that we found in Chapter 8 .

A \(450-\mathrm{g}\) mass on a spring is oscillating at \(1.2 \mathrm{~Hz}\), with total energy \(0.51 \mathrm{~J}\). What's the oscillation amplitude?

The top of a skyscraper sways back and forth, completing 95 full oscillation cycles in 10 minutes. Find (a) the period and (b) the frequency (in Hz) of its oscillatory motion.

Is the frequency of a simple harmonic motion independent of the physical configuration of the oscillating body? Explain.

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