Question

A spring is arranged vertically. When a force of 2.00 lb is applied to the free end, the spring deflects 5 inches. a) Find the spring restoration constant. Then a 0.75-lb body is hung from the free end of the spring, stretched 4 inches from the equilibrium position, and released. b) Find the oscillation frequency in Hz. c) Find the amplitude, angular frequency, phase angle and then express the position equation for any time. d) When the object has moved half between the initial position and the equilibrium position, find at that point: the kinetic energy, the potential energy and the acceleration. 

          A spring is arranged vertically. When a force of 2.00 lb is applied to the free end, the spring deflects 5 inches.
a) Find the spring restoration constant. Then a 0.75-lb body is hung from the free end of the spring, stretched 4 inches from the equilibrium position, and released.
b) Find the oscillation frequency in Hz.
c) Find the amplitude, angular frequency, phase angle and then express the position equation for any time.
d) When the object has moved half between the initial position and the equilibrium position, find at that point: the kinetic energy, the potential energy and the acceleration.
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Added by Sarah H.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A spring is arranged vertically. When a force of 2.00 lb is applied to the free end, the spring deflects 5 inches. a) Find the spring restoration constant. Then a 0.75-lb body is hung from the free end of the spring, stretched 4 inches from the equilibrium position, and released. b) Find the oscillation frequency in Hz. c) Find the amplitude, angular frequency, phase angle and then express the position equation for any time. d) When the object has moved half between the initial position and the equilibrium position, find at that point: the kinetic energy, the potential energy and the acceleration.
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A spring in a vertical position is bolted into a fixed beam. When the spring is loaded with a 120 g object, it causes the spring to be displaced from its equilibrium position by 10 cm. It is then loaded with a 150 g object and compressed upwards. Neglecting air resistance and damping, the object-spring assembly is subjected to oscillating motion. Calculate the following for an amplitude of 35 cm. a. The angular frequency in rad/s. b. The period and frequency in seconds and hertz, respectively. c. The maximum velocity in m/s. d. The maximum acceleration in m/s^2. e. The restoring force of the spring in N.

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A spring in a vertical position is bolted into a fixed beam. When the spring is loaded with a 120 g object, it causes the spring to be displaced from its equilibrium position by 10 cm. It is then loaded with a 150 g object and compressed upwards. Neglecting air resistance and damping, the object-spring assembly is subjected to oscillating motion. Calculate the following for an amplitude of 35 cm. a. The angular frequency in rad/s. b. The period and frequency in seconds and hertz, respectively. c. The maximum velocity in m/s. d. The maximum acceleration in m/s^2. e. The restoring force of the spring in N.

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Transcript

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00:01 Given the mass of the object m which is loaded first say m not is 120 gram or 0 .12 kg and for this mass the displacement is 10 centimeter that is 0 .1 meter then it is loaded with the mass equals to 150 gram that is 0 .15 kg now first we will determine the spring constant by using the equation m not g is equals to k x not this implies k is m not g upon x not so on substituting the values we get 0 .12 into 9 .8 upon 0 .1 which is equals to 11 .77 new newton per meter so this is the spring constant now, the equation of motion for the system is m d2x by d t square plus kx is equals to 0.
01:10 This implies 0 .15 d2x by d t square plus 11 .77x is equals to 0.
01:23 Let this be equation 1.
01:28 Now angular frequency is given by square root of k upon m.
01:37 So on substituting the values we get 11 .77 upon 0 .15 that is equals to 8 .85 radiance per second.
01:49 Hence this is the angular frequency.
01:54 Now period t is given by 2 pi by 5 by.
02:01 Omega so that is equals to 2 into 3 .14 upon 8 .858 equals to 0 .70925 second.
02:15 Hence this is the period and frequency is inverse of period that is frequency is 1 upon period so this will be equals to 1 upon 0 .70925 that is equals to 1 .409 hertz hence this is the frequency now maximum velocity is equals to a omega where a is the amplitude and a is given to be 35 centimeter that is equals to 0 .35 meter so on substituting the values we get 0 .35 into 8 .858 that is equals to 3 .1 meter per second...
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