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You are watching an object that is moving in SHM. When the object is displaced \(0.600 \mathrm{~m}\) to the right of its equilibrium position, it has a velocity of \(2.20 \mathrm{~m} / \mathrm{s}\) to the right and an acceleration of \(8.40 \mathrm{~m} / \mathrm{s}^{2}\) 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?

Short Answer

Expert verified
The object will move an additional 0.2887 m from the given point before it momentarily stops and starts to move back to the left.

Step by step solution

01

Understanding SHM

In SHM, we use the formula \(a=-\omega^{2}x\), where \(a\) is acceleration, \(-\omega^{2}\) is the angular frequency squared, and \(x\) is the displacement. Acceleration is directed towards the equilibrium, so it is negative when the displacement, \(x\), is positive. From the problem, we're given \(a=-8.40 \mathrm{~m} / \mathrm{s}^{2}\), and \(x=0.600 \mathrm{~m}\). Therefore, we can solve for \(\omega^{2}\) using the formula: \([-8.40 \mathrm{~m} / \mathrm{s}^{2} / 0.600 \mathrm{~m} = \omega^{2}\). This gives us \(\omega^{2}\approx -14 \mathrm{s}^{-2}\). The angular frequency is always a positive number, so simply take the absolute value: \(\omega=\sqrt{14} \mathrm{s}^{-1}\).
02

Calculating the Amplitude

To find at what displacement position the object will stop momentarily, we need to find the maximum displacement or the amplitude (A) of the oscillation, where the object’s velocity is 0. We can use the following formula derived from the energy conservation law of SHM: \(v=\omega \sqrt{A^{2}-x^{2}}\). Given are \(v=2.20 \mathrm{~m} / \mathrm{s}\), \(x=0.600 \mathrm{~m}\) and \(\omega \approx \sqrt{14} \mathrm{s}^{-1}\). Solving for \(A\), we have \(A=\sqrt{[(2.20 \mathrm{~m} / \mathrm{s}) / \sqrt{14} \mathrm{s}^{-1}]^2 + (0.600 \mathrm{~m})^{2}}\). After calculating, we find that \(A \approx 0.8887 \mathrm{m}\).
03

Finding the additional displacement

The object has already moved 0.600 m to the right. We calculate the additional displacement from this point by subtracting this current displacement from the amplitude. So, it is \( \Delta x = A - x = 0.8887 \mathrm{m} - 0.600 \mathrm{m} = 0.2887 \mathrm{m}\). The object will move an additional 0.2887 m before it momentarily stops and heads back towards equilibrium.

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Key Concepts

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

Angular Frequency
In Simple Harmonic Motion (SHM), angular frequency is a key concept that helps determine how an object oscillates over time. It is represented by the Greek letter omega (\( \omega \)). Angular frequency (\( \omega \)) is related to the frequency and period of oscillation, and it's expressed in radians per second. The equation used is \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the object.

The angular frequency also derives from the relationship in the formula \( a = -\omega^2 x \), where \( a \) is the acceleration and \( x \) represents displacement. This shows how acceleration is directly proportional to both the displacement and the square of the angular frequency. Because SHM involves periodic motion, the value of angular frequency helps us understand the rate at which the oscillation occurs.
Displacement
Displacement in SHM refers to how far the object is from its equilibrium position at any given moment. SHM assumes that the displacement is symmetric around the equilibrium point. Displacement can be positive or negative, depending on the direction from the equilibrium.

In mathematical terms, displacement in SHM is often represented as \( x(t) = A \cos(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is angular frequency, and \( \phi \) is the phase angle. This formula describes how the position of the object varies with time, showing that displacement is a function of both time and amplitude, reaching a maximum at the endpoints of its path.
Amplitude
Amplitude is a measure of the maximum extent of the motion from the equilibrium position in SHM. It's the greatest distance that the object can reach from the center point during its motion. This is symbolized by \( A \) in the equation \( x(t) = A \cos(\omega t + \phi) \).

In the context of the given problem, when the object is at an amplitude, its velocity is zero, meaning it momentarily stops before reversing direction. Amplitude also signifies the total energy in the system because energy in SHM is proportional to the square of the amplitude. Larger amplitudes equate to more energy in the system because the object travels a longer distance during each swing.
Energy Conservation in SHM
In Simple Harmonic Motion, the principle of energy conservation plays a significant role. When an object moves in SHM, its energy switches between kinetic and potential forms. Total mechanical energy remains constant if there’s no external force.
  • Kinetic Energy (KE) is highest at the equilibrium position because velocity is maximum and potential energy is zero.
  • Potential Energy (PE) is highest at the amplitude, where velocity and KE are zero.
The formula for total energy in SHM is \( E = \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 \). Using energy conservation equations helps to solve for unknowns like amplitude and velocity at different displacement points.
Velocity in SHM
Velocity in SHM describes how quickly and in what direction an object moves. Like displacement, it varies with time and oscillates back and forth. In SHM, the instantaneous velocity, \( v(t) \), is the derivative of displacement with respect to time.

The velocity function is given by \( v(t) = -A \omega \sin(\omega t + \phi) \). At the equilibrium position, the velocity is at its peak because the object moves fastest as it passes through the central point. Conversely, at points of maximum displacement (amplitude), velocity is zero since the object changes direction.

By analyzing the velocity in SHM, one can determine both the object's speed and the direction of its motion at any point in its path. Understanding velocity is crucial for predicting future motion in SHM and solving related physics problems.

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Most popular questions from this chapter

A Spring with Mass. The preceding problems in this chapter have assumed that the springs had negligible mass. But of course no spring is completely massless. To find the effect of the spring's mass, consider a spring with mass \(M,\) equilibrium length \(L_{0},\) and spring constant \(k\). When stretched or compressed to a length \(L,\) the potential energy is \(\frac{1}{2} k x^{2},\) where \(x=L-L_{0}\). (a) Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. Calculate the kinetic energy of the spring in terms of \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each piece in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L .\) The result is \(n o t \frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) (b) Take the time derivative of the conservation of energy equation, Eq. (14.21), for a mass \(m\) moving on the end of a massless spring. By comparing your results to Eq. (14.8), which defines \(\omega\), show that the angular frequency of oscillation is \(\omega=\sqrt{k / m}\). (c) Apply the procedure of part (b) to obtain the angular frequency of oscillation \(\omega\) of the spring considered in part (a). If the effective mass \(M^{\prime}\) of the spring is defined by \(\omega=\sqrt{k / M^{\prime}},\) what is \(M^{\prime}\) in terms of \(M ?\)

\(\mathrm{A} 50.0 \mathrm{~g}\) hard-boiled egg moves on the end of a spring with force constant \(k=25.0 \mathrm{~N} / \mathrm{m} .\) Its initial displacement is \(0.300 \mathrm{~m} . \mathrm{A}\) damping force \(F_{x}=-b v_{x}\) acts on the egg, and the amplitude of the motion decreases to \(0.100 \mathrm{~m}\) in \(5.00 \mathrm{~s}\). Calculate the magnitude of the damping constant \(b\).

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 friction less 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 \(15.0 \mathrm{~cm} .\) After the impact, the block moves in SHM. Calculate the period of this motion.

A slender rod of length \(80.0 \mathrm{~cm}\) and mass \(0.400 \mathrm{~kg}\) has its center of gravity at its geometrical center. But its density is not uniform; it increases by the same amount from the center of the rod out to either end. You want to determine the moment of inertia \(I_{\mathrm{cm}}\) of the rod for an axis perpendicular to the rod at its center, but you don't know its density as a function of distance along the rod, so you can't use an integration method to calculate \(I_{\mathrm{cm}}\). Therefore, you make the following measurements: You suspend the rod about an axis that is a distance \(d\) (measured in meters) above the center of the rod and measure the period \(T\) (measured in seconds) for small-amplitude oscillations about the axis. You repeat this for several values of \(d\). When you plot your data as \(T^{2}-4 \pi^{2} d / g\) versus \(1 / d\), the data lie close to a straight line that has slope \(0.320 \mathrm{~m} \cdot \mathrm{s}^{2} .\) What is the value of \(I_{\mathrm{cm}}\) for the rod?

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A\). (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium.) (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

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