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A thrill-seeking cat with mass \(4.00 \mathrm{~kg}\) is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is \(0.050 \mathrm{~m},\) 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.

Short Answer

Expert verified
At the highest point, the system posses all potential energy as the cat is momentarily at rest. At the lowest point, the system possesses all kinetic energy as the cat is in motion with maximum speed and the spring is fully stretched. At the equilibrium position, the energy is evenly distributed as both kinetic and potential energy.

Step by step solution

01

Calculate the spring constant

Firstly, we need to calculate the spring constant \(k\), which we can use the formula \[-mg = -kx\] where \(m = 4.00 kg\) is the mass of the cat, \(g = 9.81 m/s^2\) is the acceleration due to gravity and \(x = 0.050 m\) is the amplitude. Solve for \(k\) to get \(k = mg / x\).
02

Calculate the elastic potential energy at the highest point (a)

After obtaining \(k\), we can then find the elastic potential energy at the highest point of the motion using the formula \[U_{spring} = (1/2)kx^2\] where \(x = 0\) at the highest point. Hence, the spring has no potential energy at the highest point, \(U_{spring} = 0 J\).
03

Calculate the kinetic energy at the highest point (a)

We can calculate the kinetic energy of the cat at its highest point using the formula \[KE = (1/2)mv^2\] where \(v = 0\) at the highest point. Hence, the kinetic energy of the cat is 0 at the highest point, \(KE = 0 J\).
04

Calculate the gravitational potential energy at the highest point (a)

The gravitational potential energy at the highest point can be calculated using the formula \[U_{gravity} = mgh\] where \(h = 0.050 m\) is the amplitude. Solve for \(U_{gravity}\).
05

Calculate the total energy at the highest point (a)

Now, at the highest point (a), the total energy will be the sum of the gravitational potential energy, the spring potential energy and the kinetic energy. Since both the kinetic energy and the spring energy are zero, the total energy at point (a) will be the same as the gravitational potential energy found in step 4.
06

Calculate the energies at point (b) and point (c)

Using the same formulae, you can then calculate the elastic potential energy, the kinetic energy, the gravitational potential energy and the total energy at its lowest point (b) and at its equilibrium position (c). Note that at the lowest point (b), the spring is fully stretched and at the equilibrium position (c), the spring is neither stretching nor compressing so its length is at the natural unstretched length.

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Key Concepts

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

Elastic Potential Energy
Elastic potential energy is the energy stored in an object when it is compressed or stretched. It's similar to winding up a spring toy; when you let go, the stored energy is converted to motion as it springs back to its original shape.

In the context of the thrill-seeking cat attached to the spring, the elastic potential energy would be highest when the spring is at its maximum stretch or compression. You can determine the amount of this energy using the formula \(U_{spring} = \frac{1}{2}kx^2\), where \(k\) represents the spring constant and \(x\) represents the displacement of the spring from its equilibrium position. Notice that when the spring is neither stretched nor compressed, as when the cat is at its highest point, the elastic potential energy is zero, since \(x = 0\).
Kinetic Energy
Kinetic energy, on the other hand, is the energy that an object has due to its motion. Think of it as the energy you have while running — the faster you go, the more kinetic energy you have.

For the cat oscillating on the spring, the kinetic energy can be found using the formula \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the cat and \(v\) is the velocity of the cat. It’s crucial to remember that kinetic energy is zero when the cat's speed is zero, which happens at the highest and lowest points in its motion because those points are where the cat momentarily stops before changing direction.
Gravitational Potential Energy
Gravitational potential energy is the energy held by an object because of its position relative to a gravitational field, usually the Earth's gravity. It's like being at the top of a slide - you have more potential energy when you're higher up, and that energy can convert into motion as you slide down.

For our cat secured to a spring, the gravitational potential energy can be calculated with the formula \(U_{gravity} = mgh\), where \(h\) is the vertical height from the lowest point of the motion. This energy is highest when the cat is at its highest point in the spring's oscillation. In contrast, at the equilibrium position, the cat would have less gravitational potential energy since it wouldn’t be lifted above the lowest point.
Spring Constant
The spring constant, commonly symbolized as \(k\), is a measure of a spring's stiffness. Imagine using different springs on a trampoline – a spring with a higher spring constant would give you a firmer bounce, while a lower spring constant would result in a softer bounce.

In the solution to our cat's SHM problem, the spring constant is found using the relationship between the force exerted by gravity on the cat and the spring's ability to resist that force when stretched by a certain amount. The formula, \(k = \frac{mg}{x}\), links the spring constant to the cat’s weight (mass times the acceleration due to gravity) and the amplitude of the spring oscillation. With a known spring constant, we can predict how the spring will behave when supporting the cat through various points of its motion.

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Most popular questions from this chapter

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?

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is \(0.165 \mathrm{~m}\). The maximum speed of the block is \(3.90 \mathrm{~m} / \mathrm{s}\). What is the maximum magnitude of the acceleration of the block?

Object \(A\) has mass \(m_{A}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{A} .\) Object \(B\) has mass \(m_{B}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{B}\). The amplitude \(A_{A}\) for object \(A\) is twice the amplitude \(A_{B}\) for the motion of object \(B\). Also, \(m_{B}=4 m_{A}\) and \(k_{A}=9 k_{B}\). (a) What is the ratio of the maximum speeds of the two objects, \(v_{\max , A} / v_{\max , B} ?\) (b) What is the ratio of their maximum accelerations, \(a_{\max , A} / a_{\max , B} ?\)

BIO (a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note that is sung. If someone sings the note \(\mathrm{B}\) flat, which has a frequency of \(466 \mathrm{~Hz}\), how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angular frequency of the cords? (b) Hearing. When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that young humans can hear has a period of \(50.0 \mu \mathrm{s}\). What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) Vision. When light having vibrations with angular frequency ranging from \(2.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) to \(4.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) Ultrasound. High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect small objects such as tumors, a frequency of around \(5.0 \mathrm{MHz}\) is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?

Two uniform solid spheres, each with mass \(M=0.800 \mathrm{~kg}\) and radius \(R=0.0800 \mathrm{~m},\) are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k=160 \mathrm{~N} / \mathrm{m}\) has one end attached to the wall and the other end attached to a friction less ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

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