1- A 40-kg object on a frictionless horizontal surface oscillating on a spring t
ID: 1405840 • Letter: 1
Question
1- A 40-kg object on a frictionless horizontal surface oscillating on a spring that has a force constant equal to 1.5 kN/m has a total mechanical energy of 0.60 J.
(a) What is the amplitude of the motion?
....... cm
(b) What is the maximum speed?
...... cm/s
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2- Find the total energy of a 4.0 kg object oscillating on a horizontal spring with an amplitude of 10 cm and a frequency of 2.0 Hz.
....... J
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3-The period of an oscillating particle is 48 s, and its amplitude is 12 cm. At t = 0, it is at its equilibrium position. Find the distances traveled during these intervals.
(a) t = 0 to t = 12 s
....... cm
(b) t = 12 s to t = 24 s
....... cm
(c) t = 0 to t = 6 s
........ cm
(d) t = 6 s to t = 12 s
...... cm
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4- An oscillator has a period of 2.9 s. Its amplitude decreases by 8% during each cycle.
(a) By how much does its energy decrease during each cycle?
.......... %
(b) What is the time constant ?
..........) s
(c) What is the Q factor?
........
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5- An object of unknown mass is hung on the end of an unstretched spring and is released from rest. If the object falls 2.93 cm before first coming to rest, find the period of the motion.
......... s
Explanation / Answer
a)
At maximum stress/ Compression position we have maximum amplitude.
Total Energy at that point is because of potential Energy -
Spring Constant k = 1.5 kN/m
Et = (1/2)k A^2 = 0.6 J
A^2 = (0.6 * 2 )/ (1.5 * 10^3)
A = sqrt((0.6 * 2 )/ (1.5 * 10^3))
Amplitude A = 0.028 m
Amplitude A = 2.8 cm
b)
Maximum speed occurs when all the energy of the system is in the form of kinetic energy of the object. This will occur when there is no extension of the spring.
(1/2) m v^2 = 0.6 j
v = sqrt(0.6 * 2 / m)
v = sqrt(0.6 *2 / 40)
v = 0.173 m/s
Max Velocity v = 17.3 cm/s
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