Problem 16: In the analysis of a block-spring system undergoing simple harmonic motion, the mass of the spring is usually considered negligible. However, when we do account for the mass of the spring, we find that its effect is to increase the period. Let the masses of the block and spring be M and m, respectively. Assume the spring is uniform and the speed of a slice of the spring is directly proportional to its distance from the fixed end. For example, the speeds of the spring at its fixed end, its mid-point, and the point where it connects to the block are 0, v/2, and v, respectively, where v is the instantaneous speed of the block. Part (a) Enter an expression for the kinetic energy of the spring in terms of m and v. K= Part (b) Calculate the kinetic energy, in joules, of a m = 0.095-kg spring in an oscillating block-spring system, when the block is moving at a speed of v= 1.7 m/s. Part (c) Now write the kinetic energy of the spring as KE=0.5m², where v is the block's instantaneous speed and me is termed the effective mass of the spring. Enter an expression for me in terms of m and v. Part (d) Calculate the effective mass, in kilograms, of an m = 0.095-kg spring in an oscillating block-spring system. 4 Part (e) Assume the mass in the formula for the period of oscillation of a block-spring system can be replaced by M + me. For M = 1 kg and m = 0.095 kg by how much, in seconds, does the formula's value increase when the spring's effective mass is included in the calculation? Take a spring constant of k = 29 N/m. AT=
Problem 16: In the analysis of a block-spring system undergoing simple harmonic motion, the mass of the spring is usually considered negligible. However, when we do account for the mass of the spring, we find that its effect is to increase the period. Let the masses of the block and spring be M and m, respectively. Assume the spring is uniform and the speed of a slice of the spring is directly proportional to its distance from the fixed end. For example, the speeds of the spring at its fixed end, its mid-point, and the point where it connects to the block are 0, v/2, and v, respectively, where v is the instantaneous speed of the block. Part (a) Enter an expression for the kinetic energy of the spring in terms of m and v. K= Part (b) Calculate the kinetic energy, in joules, of a m = 0.095-kg spring in an oscillating block-spring system, when the block is moving at a speed of v= 1.7 m/s. Part (c) Now write the kinetic energy of the spring as KE=0.5m², where v is the block's instantaneous speed and me is termed the effective mass of the spring. Enter an expression for me in terms of m and v. Part (d) Calculate the effective mass, in kilograms, of an m = 0.095-kg spring in an oscillating block-spring system. 4 Part (e) Assume the mass in the formula for the period of oscillation of a block-spring system can be replaced by M + me. For M = 1 kg and m = 0.095 kg by how much, in seconds, does the formula's value increase when the spring's effective mass is included in the calculation? Take a spring constant of k = 29 N/m. AT=
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Concept explainers
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
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Transcribed Image Text:Problem 16: In the analysis of a block-spring system undergoing simple harmonic motion, the mass of the spring is usually considered negligible.
However, when we do account for the mass of the spring, we find that its effect is to increase the period. Let the masses of the block and spring be M and m,
respectively. Assume the spring is uniform and the speed of a slice of the spring is directly proportional to its distance from the fixed end. For example, the
speeds of the spring at its fixed end, its mid-point, and the point where it connects to the block are 0, v/2, and v, respectively, where v is the instantaneous
speed of the block.
Part (a) Enter an expression for the kinetic energy of the spring in terms of m and v.
K=
Part (b) Calculate the kinetic energy, in joules, of a m = 0.095-kg spring in an
oscillating block-spring system, when the block is moving at a speed of v = 1.7 m/s.
Part (c) Now write the kinetic energy of the spring as KE = 0.5m₂v², where v is the
block's instantaneous speed and me is termed the effective mass of the spring. Enter an expression
for me in terms of m and v.
Part (d) Calculate the effective mass, in kilograms, of an m = 0.095-kg spring in an
oscillating block-spring system.
Part (e) Assume the mass in the formula for the period of oscillation of a block-spring
system can be replaced by M+me. For M = 1 kg and m = 0.095 kg by how much, in seconds, does
the formula's value increase when the spring's effective mass is included in the calculation? Take
a spring constant of k = 29 N/m.
AT=
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