Table of Contents
An Oscillating Spring
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| An oscillating spring is an example of SHM (Simple Harmonic Motion) which results when the force is directly |
| proportional to the displacement and is always directed toward the equilibrium position. As a result of the restoring force |
| changing, the acceleration is also changing, and the kinematics equations for constant acceleration do not apply. |
| Using the example of a mass attached to a vertical spring as shown in the animation gives the result: |
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F = ma = -kx
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a = -k/m • x
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| where k is the spring constant with units of N/m (newtons/meter), m is the mass of the object with units of kg, and x is |
| the displacement from the equilibrium with units of m (meters). The negative sign results from the restoring force always |
| opposing the motion and being directed toward the equilibrium position. The second equation demonstrates that the |
| acceleration at each instant is directly proportional to the negative displacement at that instant. |
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| This situation is analagous to that of a simple pendulum. When x is a maximum at the amplitude, the restoring force, Fr, |
| and the resulting acceleration is also a maximum. At this point, the velocity is zero. When x is zero at the equilibrium |
| position, both Fr and a are zero with the velocity a maximum. |
| Analagous to a pendulum, one can apply the conservation of mechanical energy to the mass-spring system. |
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ΔE = 0
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(KE + PEe)eq = (KE + PEe)A
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| where eq signifies the equilibrium position and A is the amplitude. |
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KE + 0 = 0 + PEe
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1/2mv2 = 1/2kA2
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v = (kA2/m)1/2 = (k/m)1/2A
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| The following equations will be accepted without proof. |
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| Modifying the x component of displacement yields: |
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x = Asin((k/m)1/2 • t)
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| The quantity in parenthesis is the angle measured in radians. Because x is a function of time, it is said to be a sinusoidal |
| function. The sin function repeats itself when the quantity (k/m)1/2 • t increases by 2π. The period of the oscillating |
| spring is given by: |
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T = 2π(m/k)1/2
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| Examining this equation leads to four important considerations: |
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large frequency. |
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| 1) A 5.0 kg mass is attached to a vertical spring and stretches the spring 0.10 m. The spring is then pulled down |
| 0.050 m and released. Determine the: |
| (a) amplitude |
| (b) period |
| (c) frequency |
| 2) A 100. g mass hangs from a vertical spring. When it is pulled 10. cm below its equilibrium position and released, it |
| oscillates with a period of 2.0 s. |
| (a) What is the maximum velocity of the mass? |
| (b) What is the acceleration when it is 5.0 cm below the equilibrium position? |
| (c) How much will the spring shorten if the mass is removed? |
| 3) When a 0.150 kg mass is attached to a vertical spring, it stretches by 0.0400 m. The spring is then pulled down |
| 0.300 m below the equilibrium position and released. Determine: |
| (a) force constant of the spring |
| (b) period of the motion |
| (c) mass’s displacement as a function of time |
| (d) mass’s velocity as a function of time |
| (e) mass’s acceleration as a function of time |
| 4) A mass attached to a vertical spring is vibrating with a frequency of 3.0 s-1 and has an amplitude of 16 cm. Determine |
| the: |
| (a) maximum velocity and the maximum acceleration |
| (b) velocity and acceleration when the mass has a displacement of 7.0 cm |
| (c) time needed to move from the equilibrium position to a displacement of 13 cm |
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