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Chapter 14: Problem 9

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

Short Answer

Expert verified
(a) 0.150 s to go from 0.320 m to 0.160 m and (b) 0.075 s from 0.160 m to 0.

Step by step solution

01

Understanding SHM Equation

The object undergoes Simple Harmonic Motion (SHM), defined by the equation: \\[ x(t) = A \cos(\omega t + \phi) \] \Given the amplitude \(A = 0.320\,m\), the period \(T = 0.900\,s\), and that the object starts from rest at \(x = A\), we deduce \(\phi = 0\) and \(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.900} \). This gives: \\[ x(t) = 0.320 \cos\left(\frac{2\pi}{0.900} t\right) \]
02

Calculate Time to Reach x = 0.160 m

Substitute \(x = 0.160\,m\) into the SHM equation to find the time \(t\): \\[ 0.160 = 0.320 \cos\left(\frac{2\pi}{0.900} t\right) \] \Solving, this yields \( \cos\left(\frac{2\pi}{0.900} t\right) = 0.5 \). Therefore, \\[ \frac{2\pi}{0.900} t = \cos^{-1}(0.5) = \frac{\pi}{3} \] \Hence, \( t = \frac{0.900}{2\pi} \times \frac{\pi}{3} = 0.150\,s \).
03

Determine Time from x = 0.160 m to x = 0

For \(x = 0\), the equation becomes: \\[ 0 = 0.320 \cos\left(\frac{2\pi}{0.900} t\right) \] \Which implies \( \cos\left(\frac{2\pi}{0.900} t\right) = 0 \). Thus, \\[ \frac{2\pi}{0.900} t = \frac{\pi}{2} \] \Then \( t = \frac{0.900}{2\pi} \times \frac{\pi}{2} = 0.225\,s \). \Subtract the previous time (0.150 s) to find the time taken to go from \(0.160\,m\) to \(0\). \This results in \( t = 0.225 - 0.150 = 0.075\,s \).

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Key Concepts

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

Amplitude
In Simple Harmonic Motion (SHM), the amplitude ( A ) represents the maximum extent of displacement from the equilibrium position. Think of it as the furthest distance the object travels from its resting point. It is a crucial parameter in SHM since it defines the peak value of the oscillation.
  • In this context, the amplitude is given as 0.320 m, meaning that the object travels between +0.320 m and -0.320 m at extreme positions.
  • The amplitude does not change over time in ideal SHM, assuming there are no external forces like friction.
  • Understanding amplitude helps visualize the motion range of the object.
Knowing the amplitude is vital because it sets the stage for analyzing other aspects of SHM, such as energy or time calculations.
Period
The period (T) in SHM refers to the time it takes for the object to complete one full cycle of motion. It determines how quickly the object swings back and forth. A shorter period indicates a faster oscillation, while a longer period implies slower motion.
  • In our scenario, the period is 0.900 seconds. This tells us the object goes from start to peak and back to the start in 0.900 seconds.
  • The period is inversely linked to the angular frequency (\( \omega \)), calculated by (\( \omega = \frac{2\pi}{T} \)).
  • Knowing the period allows us to determine the timing of the object's motion at any point.
Understanding the period helps establish the rhythm of the oscillation and is necessary for time-based problems, like determining the time to reach certain positions.
SHM Equation
The SHM equation is a mathematical representation that describes the motion of an oscillating object. It typically takes the form:
\[ x(t) = A \cos(\omega t + \phi) \]
This equation helps predict the position of the object (x(t)) at any given time (t).
  • Given the amplitude (\( A \)) and period, we can determine (\( \omega \)), angular frequency, as \( \omega = \frac{2\pi}{T} \).
  • The phase constant (\( \phi \)) accounts for the starting position; here, it is 0 since the object starts at its maximum displacement.
  • The equation combines these variables to model the motion accurately.
Utilizing the SHM equation is essential for calculating specific times or positions during the motion, providing a comprehensive method to understand and predict the behavior of oscillating systems.
Cosine Function
The cosine function is crucial in SHM as it models the oscillation pattern. This periodic function, represented by (\( \cos \)), oscillates between -1 and 1.
  • In the context of SHM, the cosine function is used within the equation (\( x(t) = A \cos(\omega t + \phi) \)) to describe the repetitive motion.
  • The use of the cosine function ensures the object's position toggles smoothly between its limits, exactly like a swinging pendulum or a plucked guitar string.
  • Since it originates from trigonometry, it provides a precise way to represent oscillations over time, with the wave-like nature mimicking physical movement.
The cosine function's characteristics of smooth, continuous cycles make it ideal for modeling periodic SHM, capturing the essence of back-and-forth motion in a mathematically sound manner.

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Most popular questions from this chapter

CP CALC For a certain oscillator the net force on the body with mass \(m\) is given by \(F_{x}=-c x^{3}\) (a) What is the potential energy function for this oscillator if we take \(U=0\) at \(x=0\) ? (b) One-quarter of a period is the time for the body to move from \(x=0\) to \(x=A .\) Calculate this time and hence the period. [Hint: Begin with \(\mathrm{Eq} .(14.20),\) modified to include the potential-energy function you found in part (a), and solve for the velocity \(v_{x}\) as a function of \(x\) . Then replace \(v_{x}\) with \(d x / d t\) . Separate the variable by writing all factors containing \(x\) on one side and all factors containing \(t\) on the other side so that each side can be integrated. In the \(x\) -integral make the change of variable \(u=x / A .\) The resulting integral can be evaluated by numerical methods on a computer and has the value \(\int_{0}^{1} d u / \sqrt{1-u^{4}}=1.31 .1(\mathrm{c})\) According to the result you obtained in part (b), does the period depend on the amplitude A of the motion? Are the oscillations simple harmonic?

You pull a simple pendulum 0.240 m long to the side through an angle of \(3.50^{\circ}\) 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 \(1.75^{\circ}\) instead of \(3.50^{\circ} ?\)

A 5.00 -kg partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.20 s. (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is 0.050 \(\mathrm{m}\) above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point 0.050 \(\mathrm{m}\) below its equilibrium position to a point 0.050 \(\mathrm{m}\) above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

A 1.50 -kg mass on a spring has displacement as a function of time given by the equation $$x(t)=(7.40 \mathrm{cm}) \cos \left[\left(4.16 \mathrm{s}^{-1}\right) t-2.42\right]$$ 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 \(t=1.00 \mathrm{s} ;\) (f) the force on the mass at that time.

A 1.80 -kg connecting rod from a car engine is pivoted about a horizontal knife edge as shown in Fig. E14.53. The center of gravity of the rod was located by balancing and is 0.200 \(\mathrm{m}\) from the pivot. When the rod is set into small-amplitude oscillation, it makes 100 complete swings in 120 s. Calculate the moment of inertia of the rod about the rotation axis through the pivot.
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