
Concept explainers
(a)
The relation among the six lengths.
(a)

Answer to Problem 11.1P
The relation between the lengths is
Explanation of Solution
In the given case,
The net force acting on each of the springs is same in equilibrium position and each of the springs experiences either tension of compression. The restoring force of the spring is directly proportional to the compressed length according to Hooke’s law.
Here,
Rewrite equation (I) for the force acting on the first spring.
Here,
The compressed length of the first spring can be written as
Here,
Substituting equation (III) in equation (II).
Similarly, the equation for the restoring force on the second spring can be written as
Here,
And the equation for the restoring force on the third springs is
Here,
Conclusion:
The forcing acting on each of the spring is same at equilibrium. Hence equate equation (IV), (V) and (VI).
Therefore, the relation between the lengths is
(b)
The validity of the equation.
(b)

Answer to Problem 11.1P
The relations are proved.
Explanation of Solution
Two different forces act on the first cart. The extensive force of the first spring and the compressive force of the second spring.
Write the equation for the extensive force on the first cart.
Here,
Write the equation for the compressive force acting on the first cart by the second spring is,
Here,
Write the equation for the net force acting on the first cart suing Newton’s second law of motion.
Here,
The net force acting on the first cart is the vector sum of the forces
Substitute equation (VIII), (VII) and (VI) in equation (IX)
Consider the case of the case of the second cart. The forces acting on the second cart is the extensive force of the second spring and compressive force of the third spring.
Write the equation for the compressive acting on the second cart is,
Here,
Write the equation for the extensive force acting on the first cart by the second spring.
Here,
Write the equation for the net force acting on the second cart suing Newton’s second law of motion.
Here,
The net force acting on the second cart is
The negative sign indicates that the force
Substitute equation (XIII), (XII) and (XI) in equation (XIV).
Conclusion:
Therefore, the net force on first cart is
Want to see more full solutions like this?
Chapter 11 Solutions
Classical Mechanics
- For number 11 please sketch the harmonic on graphing paper.arrow_forward# E 94 20 13. Time a) What is the frequency of the above wave? b) What is the period? c) Highlight the second cycle d) Sketch the sine wave of the second harmonic of this wave % 7 & 5 6 7 8 * ∞ Y U 9 0 0 P 150arrow_forwardShow work using graphing paperarrow_forward
- Can someone help me answer this physics 2 questions. Thank you.arrow_forwardFour capacitors are connected as shown in the figure below. (Let C = 12.0 μF.) a C 3.00 με Hh. 6.00 με 20.0 με HE (a) Find the equivalent capacitance between points a and b. 5.92 HF (b) Calculate the charge on each capacitor, taking AV ab = 16.0 V. 20.0 uF capacitor 94.7 6.00 uF capacitor 67.6 32.14 3.00 µF capacitor capacitor C ☑ με με The 3 µF and 12.0 uF capacitors are in series and that combination is in parallel with the 6 μF capacitor. What quantity is the same for capacitors in parallel? μC 32.14 ☑ You are correct that the charge on this capacitor will be the same as the charge on the 3 μF capacitor. μCarrow_forwardIn the pivot assignment, we observed waves moving on a string stretched by hanging weights. We noticed that certain frequencies produced standing waves. One such situation is shown below: 0 ст Direct Measurement ©2015 Peter Bohacek I. 20 0 cm 10 20 30 40 50 60 70 80 90 100 Which Harmonic is this? Do NOT include units! What is the wavelength of this wave in cm with only no decimal places? If the speed of this wave is 2500 cm/s, what is the frequency of this harmonic (in Hz, with NO decimal places)?arrow_forward
- Four capacitors are connected as shown in the figure below. (Let C = 12.0 µF.) A circuit consists of four capacitors. It begins at point a before the wire splits in two directions. On the upper split, there is a capacitor C followed by a 3.00 µF capacitor. On the lower split, there is a 6.00 µF capacitor. The two splits reconnect and are followed by a 20.0 µF capacitor, which is then followed by point b. (a) Find the equivalent capacitance between points a and b. µF(b) Calculate the charge on each capacitor, taking ΔVab = 16.0 V. 20.0 µF capacitor µC 6.00 µF capacitor µC 3.00 µF capacitor µC capacitor C µCarrow_forwardTwo conductors having net charges of +14.0 µC and -14.0 µC have a potential difference of 14.0 V between them. (a) Determine the capacitance of the system. F (b) What is the potential difference between the two conductors if the charges on each are increased to +196.0 µC and -196.0 µC? Varrow_forwardPlease see the attached image and answer the set of questions with proof.arrow_forward
- How, Please type the whole transcript correctly using comma and periods as needed. I have uploaded the picture of a video on YouTube. Thanks,arrow_forwardA spectra is a graph that has amplitude on the Y-axis and frequency on the X-axis. A harmonic spectra simply draws a vertical line at each frequency that a harmonic would be produced. The height of the line indicates the amplitude at which that harmonic would be produced. If the Fo of a sound is 125 Hz, please sketch a spectra (amplitude on the Y axis, frequency on the X axis) of the harmonic series up to the 4th harmonic. Include actual values on Y and X axis.arrow_forwardSketch a sign wave depicting 3 seconds of wave activity for a 5 Hz tone.arrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON





