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If you are interested in this interaction, find the force between the molecules by taking the derivative of the potential energy function. You will see immediately that the force does not resemble a Hooke’s law force $\left(F=\text{\u2212}kx\right)$ , but if you are familiar with the binomial theorem:
the force can be approximated by a Hooke’s law force.
Getting back to the system of a block and a spring in [link] , once the block is released from rest, it begins to move in the negative direction toward the equilibrium position. The potential energy decreases and the magnitude of the velocity and the kinetic energy increase. At time $t=T\text{/}4$ , the block reaches the equilibrium position $x=0.00\phantom{\rule{0.2em}{0ex}}\text{m,}$ where the force on the block and the potential energy are zero. At the equilibrium position, the block reaches a negative velocity with a magnitude equal to the maximum velocity $v=\text{\u2212}A\omega $ . The kinetic energy is maximum and equal to $K=\frac{1}{2}m{v}^{2}=\frac{1}{2}m{A}^{2}{\omega}^{2}=\frac{1}{2}k{A}^{2}.$ At this point, the force on the block is zero, but momentum carries the block, and it continues in the negative direction toward $x=\text{\u2212}A$ . As the block continues to move, the force on it acts in the positive direction and the magnitude of the velocity and kinetic energy decrease. The potential energy increases as the spring compresses. At time $t=T\text{/}2$ , the block reaches $x=\text{\u2212}A$ . Here the velocity and kinetic energy are equal to zero. The force on the block is $F=+kA$ and the potential energy stored in the spring is $U=\frac{1}{2}k{A}^{2}$ . During the oscillations, the total energy is constant and equal to the sum of the potential energy and the kinetic energy of the system,
The equation for the energy associated with SHM can be solved to find the magnitude of the velocity at any position:
The energy in a simple harmonic oscillator is proportional to the square of the amplitude. When considering many forms of oscillations, you will find the energy proportional to the amplitude squared.
Check Your Understanding Why would it hurt more if you snapped your hand with a ruler than with a loose spring, even if the displacement of each system is equal?
The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater force, which hurts more.
Check Your Understanding Identify one way you could decrease the maximum velocity of a simple harmonic oscillator.
You could increase the mass of the object that is oscillating. Other options would be to reduce the amplitude, or use a less stiff spring.
Describe a system in which elastic potential energy is stored.
In a car, elastic potential energy is stored when the shock is extended or compressed. In some running shoes elastic potential energy is stored in the compression of the material of the soles of the running shoes. In pole vaulting, elastic potential energy is stored in the bending of the pole.
Explain in terms of energy how dissipative forces such as friction reduce the amplitude of a harmonic oscillator. Also explain how a driving mechanism can compensate. (A pendulum clock is such a system.)
The temperature of the atmosphere oscillates from a maximum near noontime and a minimum near sunrise. Would you consider the atmosphere to be in stable or unstable equilibrium?
The overall system is stable. There may be times when the stability is interrupted by a storm, but the driving force provided by the sun bring the atmosphere back into a stable pattern.
Fish are hung on a spring scale to determine their mass. (a) What is the force constant of the spring in such a scale if it the spring stretches 8.00 cm for a 10.0 kg load? (b) What is the mass of a fish that stretches the spring 5.50 cm? (c) How far apart are the half-kilogram marks on the scale?
It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective force constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85-kg team?
a. $1.57\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{N/m}$ ; b. 77 kg, yes, he is eligible to play
One type of BB gun uses a spring-driven plunger to blow the BB from its barrel. (a) Calculate the force constant of its plunger’s spring if you must compress it 0.150 m to drive the 0.0500-kg plunger to a top speed of 20.0 m/s. (b) What force must be exerted to compress the spring?
When an 80.0-kg man stands on a pogo stick, the spring is compressed 0.120 m. (a) What is the force constant of the spring? (b) Will the spring be compressed more when he hops down the road?
a. $6.53\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N/m}$ ; b. yes, when the man is at his lowest point in his hopping the spring will be compressed the most
A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it. (a) What is the force constant of the spring? (b) What is the unloaded length of the spring?
The length of nylon rope from which a mountain climber is suspended has an effective force constant of $1.40\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{N/m}$ . (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack? ( Hint: Use conservation of energy.) (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.
a. 1.99 Hz; b. 50.2 cm; c. 0.710 m
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