Concept explainers
(a)
The total energy of the oscillating system.
(a)
Answer to Problem 27P
The total energy of the oscillating system is
Explanation of Solution
Given that the force constant of the spring is
Write the expression for the energy of the spring-object oscillating system.
Here,
Conclusion:
Substitute
Therefore, the total energy of the oscillating system is
(b)
The speed of the object when its position is
(b)
Answer to Problem 27P
The speed of the object when its position is
Explanation of Solution
Given that the force constant of the spring is
Write the expression for the speed at a given position of an object executing
Here,
Write the expression for the angular frequency of the spring-object system.
Use equation (III) in (II).
Conclusion:
Substitute
Therefore, the speed of the object when its position is
(c)
The kinetic energy of the object when its position is
(c)
Answer to Problem 27P
The kinetic energy of the object when its position is
Explanation of Solution
Write the expression for the kinetic energy of the oscillating object.
Here,
Equation (I) gives the total energy of the system.
Write the expression for the potential energy of the object at the given position.
Use equation (I) and (VI) in (V).
Conclusion:
Substitute
Therefore, the kinetic energy of the object when its position is
(d)
The potential energy of the object when its position is
(d)
Answer to Problem 27P
The potential energy of the object when its position is
Explanation of Solution
It is obtained that the total energy of the system is
Write the expression for the potential energy of the object.
Conclusion:
Substitute
Therefore, the potential energy of the object when its position is
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Chapter 15 Solutions
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
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- A grandfather clock has a pendulum length of 0.7 m and mass bob of 0.4 kg. A mass of 2 kg falls 0.8 m in seven days to keep the amplitude (from equilibrium) of the pendulum oscillation steady at 0.03 rad. What is the Q of the system?arrow_forwardUse the data in Table P16.59 for a block of mass m = 0.250 kg and assume friction is negligible. a. Write an expression for the force FH exerted by the spring on the block. b. Sketch FH versus t.arrow_forwardA lightweight spring with spring constant k = 225 N/m is attached to a block of mass m1 = 4.50 kg on a frictionless, horizontal table. The blockspring system is initially in the equilibrium configuration. A second block of mass m2 = 3.00 kg is then pushed against the first block, compressing the spring by x = 15.0 cm as in Figure P16.77A. When the force on the second block is removed, the spring pushes both blocks to the right. The block m2 loses contact with the springblock 1 system when the blocks reach the equilibrium configuration of the spring (Fig. P16.77B). a. What is the subsequent speed of block 2? b. Compare the speed of block 1 when it again passes through the equilibrium position with the speed of block 2 found in part (a). 77. (a) The energy of the system initially is entirely potential energy. E0=U0=12kymax2=12(225N/m)(0.150m)2=2.53J At the equilibrium position, the total energy is the total kinetic energy of both blocks: 12(m1+m2)v2=12(4.50kg+3.00kg)v2=(3.75kg)v2=2.53J Therefore, the speed of each block is v=2.53J3.75kg=0.822m/s (b) Once the second block loses contact, the first block is moving at the speed found in part (a) at the equilibrium position. The energy 01 this spring-block 1 system is conserved, so when it returns to the equilibrium position, it will be traveling at the same speed in the opposite direction, or v=0.822m/s. FIGURE P16.77arrow_forward
- Review. A 0.250-kg block resting on a frictionless, horizontal surface is attached to a spring whose force constant is 83.8 N/m as in Figure P15.15. A horizontal force F causes the spring to stretch a distance of 5.46 cm from its equilibrium position. (a) Find the magnitude of F. (b) What is the total energy stored in the system when the spring is stretched? (c) Find the magnitude of the acceleration of the block just after the applied force is removed. (d) Find the speed of the block when it first reaches the equilibrium position. (e) If the surface is not frictionless but the block still reaches the equilibrium position, would your answer to part (d) be larger or smaller? (f) What other information would you need to know to find the actual answer to part (d) in this case? (g) What is the largest value of the coefficient of friction that would allow the block to reach the equilibrium position? Figure P15.15arrow_forwardThe total energy of a simple harmonic oscillator with amplitude 3.00 cm is 0.500 J. a. What is the kinetic energy of the system when the position of the oscillator is 0.750 cm? b. What is the potential energy of the system at this position? c. What is the position for which the potential energy of the system is equal to its kinetic energy? d. For a simple harmonic oscillator, what, if any, are the positions for which the kinetic energy of the system exceeds the maximum potential energy of the system? Explain your answer. FIGURE P16.73arrow_forwardThe position of a particle attached to a vertical spring is given by y=(y0cost)j. The y axis points upward, y0 = 14.5 cm. and = 18.85 rad/s. Find the position of the particle at a. t = 0 and b. t = 9.0 s. Give your answers in centimeters.arrow_forward
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