PRACTICE IT Use the worked example above to help you solve this problem. A 0.545 kg object connected to a light spring with a spring constant of 19.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. Vmax = 0.177 X Your response is off by a multiple of ten. m/s (b) What is the velocity of the object when the displacement is 2.00 cm? + m/s (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE= PE= J EXERCISE For what values of x is the speed of the object 0.14 m/s? x = ± 2.0 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. cm HINTS: GETTING STARTED | I'M STUCK!
PRACTICE IT Use the worked example above to help you solve this problem. A 0.545 kg object connected to a light spring with a spring constant of 19.0 N/m oscillates on a frictionless horizontal surface. (a) Calculate the total energy of the system and the maximum speed of the object if the amplitude of the motion is 3.00 cm. Vmax = 0.177 X Your response is off by a multiple of ten. m/s (b) What is the velocity of the object when the displacement is 2.00 cm? + m/s (c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm. KE= PE= J EXERCISE For what values of x is the speed of the object 0.14 m/s? x = ± 2.0 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. cm HINTS: GETTING STARTED | I'M STUCK!
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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.
Topic Video
Question
![SOLUTION
(A) Calculate the total energy and maximum speed if the amplitude is 3.00 cm.
E = KE + PEg + PES
= 0 + 0 + 1/2k4² = 2(20.0 N/m) (3.00 x 10-² m)²
= 9.00 x 10-³ J
Substitute x = A= 3.00 cm and
k = 20.0 N/m into the equation for the
total mechanical energy E.
Use conservation of energy with x = A
and xy = 0 to compute the speed of the
object at the origin.
Substitute known values.
(KE + PEg + PE§); = (KE + PEg + PE§)ƒ
0 + 0 + 1/2kA² = 1/2mv2max + 0 + 0
1/2mv2 max = 9.00 x 10-³ J
Substitute into the equation for spring
potential energy.
"max =
(B) Compute the velocity of the object when the displacement is 2.00 cm.
v = ± √k/m (A²- - x²)
20.0 N/m
0.500 kg
LEARN MORE
1.80 x 10-2 J
0.500 kg
v = t
= 0.190 m/s
= ±0.141 m/s
(C) Compute the kinetic and potential energies when the displacement is 2.00 cm.
Substitute into the equation for kinetic
energy.
[(0.0300 m)² (0.0200 m
½
1²]) "2²
KE = 1/2mv² = 1/2(0.500 kg)(0.141 m/s)² = 4.97 x 10-³ J
PE = 1/2kx2 = 1/2 (20.0 N/m) (2.00 x 10-² m)²
= 4.00 x 10-3 J](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F93cc824a-3a30-466f-aebc-439916aac679%2Fb8f7c16e-1b3d-4d6d-be2e-f700662fc46f%2Ftvf6mt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:SOLUTION
(A) Calculate the total energy and maximum speed if the amplitude is 3.00 cm.
E = KE + PEg + PES
= 0 + 0 + 1/2k4² = 2(20.0 N/m) (3.00 x 10-² m)²
= 9.00 x 10-³ J
Substitute x = A= 3.00 cm and
k = 20.0 N/m into the equation for the
total mechanical energy E.
Use conservation of energy with x = A
and xy = 0 to compute the speed of the
object at the origin.
Substitute known values.
(KE + PEg + PE§); = (KE + PEg + PE§)ƒ
0 + 0 + 1/2kA² = 1/2mv2max + 0 + 0
1/2mv2 max = 9.00 x 10-³ J
Substitute into the equation for spring
potential energy.
"max =
(B) Compute the velocity of the object when the displacement is 2.00 cm.
v = ± √k/m (A²- - x²)
20.0 N/m
0.500 kg
LEARN MORE
1.80 x 10-2 J
0.500 kg
v = t
= 0.190 m/s
= ±0.141 m/s
(C) Compute the kinetic and potential energies when the displacement is 2.00 cm.
Substitute into the equation for kinetic
energy.
[(0.0300 m)² (0.0200 m
½
1²]) "2²
KE = 1/2mv² = 1/2(0.500 kg)(0.141 m/s)² = 4.97 x 10-³ J
PE = 1/2kx2 = 1/2 (20.0 N/m) (2.00 x 10-² m)²
= 4.00 x 10-3 J
Transcribed Image Text:PRACTICE IT
Use the worked example above to help you solve this problem. A 0.545 kg object connected to a light
spring with a spring constant of 19.0 N/m oscillates on a frictionless horizontal surface.
(a) Calculate the total energy of the system and the maximum speed of the object if the
amplitude of the motion is 3.00 cm.
E =
Vmax =
0.177 X
Your response is off by a multiple of ten. m/s
(b) What is the velocity of the object when the displacement is 2.00 cm?
±
m/s
(c) Compute the kinetic and potential energies of the system when the displacement is 2.00 cm.
KE =
PE=
EXERCISE
For what values of x is the speed of the object 0.14 m/s?
x = ± 2.0
X
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a
mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize
roundoff error. cm
HINTS:
GETTING STARTED I I'M STUCK!
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