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Chapter 14: Problem 11

An object is undergoing \(\textbf{SHM}\) with period 0.900 s and amplitude 0.320 m. At \(t =\) 0 the object is at \(x =\) 0.320 m and is instantaneously at rest. Calculate the time it takes the object to go (a) from \(x =\) 0.320 m to \(x =\) 0.160 m and (b) from \(x =\) 0.160 m to \(x =\) 0.

Short Answer

Expert verified
Time from 0.320 m to 0.160 m is 0.150 s; from 0.160 m to 0 is 0.075 s.

Step by step solution

01

Understanding the Parameters of SHM

Simple Harmonic Motion (SHM) can be described by the equation \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. Given that the amplitude \( A = 0.320 \text{ m} \) and the object starts at the maximum amplitude, \( \phi = 0 \). Thus, \( x(t) = 0.320 \cos(\omega t) \).
02

Calculating Angular Frequency

The angular frequency \( \omega \) is related to the period \( T \) by \( \omega = \frac{2\pi}{T} \). Given \( T = 0.900 \text{ s} \), we find \( \omega = \frac{2\pi}{0.900} \approx 6.98 \text{ rad/s} \).
03

Finding Time from \( x = 0.320 \text{ m} \) to \( x = 0.160 \text{ m} \)

To find the time \( t_1 \) it takes to move from \( x = 0.320 \text{ m} \) to \( x = 0.160 \text{ m} \), we use the equation \( 0.160 = 0.320 \cos(6.98 t_1) \). Solving for \( t_1 \), we have \( \cos(6.98 t_1) = 0.5 \), which implies \( 6.98 t_1 = \cos^{-1} 0.5 = \frac{\pi}{3} \). Thus, \( t_1 = \frac{\pi}{3 \times 6.98} \approx 0.150 \text{ s} \).
04

Calculating Time from \( x = 0.160 \text{ m} \) to \( x = 0 \text{ m} \)

The additional time \( t_2 \) to go from \( x = 0.160 \text{ m} \) to \( x = 0 \) can be found by solving \( 0 = 0.320 \cos(6.98 (t_1 + t_2)) \). This empties at \( \cos(6.98 (t_1 + t_2)) = 0 \), which implies \( 6.98 (t_1 + t_2) = \frac{\pi}{2} \) or alternatively any suitable increment of \( \pi/2 \). For the first occurrence merely \( 6.98 t_2 = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} \). Solving for \( t_2 \), we find \( t_2 = \frac{\pi}{6 \times 6.98} \approx 0.075 \text{ s} \).

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Key Concepts

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

Angular Frequency
Angular frequency, represented by \( \omega \), is a crucial concept in understanding Simple Harmonic Motion (SHM). It tells us how fast the oscillations happen. Think of it as the speed at which the object moves through its cycle.

Mathematically, angular frequency is related to the period \( T \) of the motion, with the formula:\[\omega = \frac{2\pi}{T}\]Here, \( 2\pi \) represents a full circle in radians, as every oscillation is essentially a circular movement in motion terms.

In our example, the period \( T \) is given as 0.900 seconds. By substituting into the formula, we calculate the angular frequency as approximately 6.98 rad/s.
  • The larger the angular frequency, the faster the object oscillates.
  • It is measured in radians per second (rad/s).
Understanding \( \omega \) helps you predict how quickly or slowly an object in SHM moves across its path.
Amplitude
In the context of Simple Harmonic Motion, amplitude denoted as \( A \), represents the maximum displacement of the oscillating object from its central or equilibrium position. Imagine it as the height of a swing; the higher you go, the larger the amplitude.

In our given problem, the amplitude is 0.320 meters. This tells us the furthest distance the object can be from its starting neutral point.
  • Amplitude is a constant value in SHM, meaning it does not change over time for the oscillation unless acted on by an external force.
  • It defines the intensity or strength of the motion.
Understanding amplitude is crucial because it gives you insights into the energy of the system. Higher amplitudes generally mean more energy is involved in the oscillations.
Phase Constant
The phase constant, represented as \( \phi \), plays a subtle but important role in SHM equations. It determines where in its cycle the object begins its motion.

In our scenario, since the object starts at the maximum amplitude when \( t = 0 \), we understand that the phase constant \( \phi \) is 0. This means the object begins oscillating from the maximum point on its path.
  • \( \phi \) can alter the position of the object at \( t = 0 \).
  • It's measured in radians, similar to angles in geometry.
Knowing \( \phi \) is vital for accurately predicting the initial condition of the object at any given time, making it easier to map out its movement on the SHM path.
Period of Oscillation
The period of oscillation, denoted by \( T \), describes the time required for the object to complete one full cycle of its motion before repeating itself. It's like measuring how long it takes for a pendulum to swing back to its starting point.

For our example, \( T = 0.900 \) seconds indicates that in less than one second, the object completes a full oscillation back to its starting position.
  • The period is directly related to the frequency of oscillation.
  • A longer period means slower oscillation, while a shorter period signifies faster oscillation.
Understanding the period helps you anticipate when the oscillation will return to a specific point in its cycle, a vital part of predicting and analyzing SHM behavior.

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Most popular questions from this chapter

A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; (b) at its lowest point; (c) at its equilibrium position.

A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A\), which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (\(Hint\): This does \(not\) occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

A 1.50-kg mass on a spring has displacement as a function of time given by $$x(t) = 7.40 \mathrm{cm}) \mathrm{cos} [ (4.16 \mathrm{rad}/s)t - 2.42] $$ 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 rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 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 15.0 cm. After the impact, the block moves in SHM. Calculate the period of this motion.

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 m/s to the right and an acceleration of 8.40 m/s\(^2\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?
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