Concept explainers
(a)
The position function of the object.
Answer to Problem 45QAP
The position of the object is
Explanation of Solution
Introduction:
The displacement of the spring performing motion is given by,
Where,
t = time
The displacement of the spring performing motion is given by,
But,
Conclusion:
The position of the object is
(b)
The velocity of the object at
Answer to Problem 45QAP
The velocity of the object at
Explanation of Solution
Introduction:
The velocity of the particle performing
Where,
t = time
Calculation:
The velocity of the particle performing simple harmonic motion,
Conclusion:
The velocity of the object at
(c)
The acceleration of the object at
Answer to Problem 45QAP
The acceleration of the object at
Explanation of Solution
Introduction:
The acceleration of the particle performing simple harmonic motion,
Where,
t = time
Calculation:
The acceleration of the particle performing simple harmonic motion,
Conclusion:
The acceleration of the object at
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Chapter 12 Solutions
COLLEGE PHYSICS,VOLUME 1
- Use 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_forwardRefer to the problem of the two coupled oscillators discussed in Section 12.2. Show that the total energy of the system is constant. (Calculate the kinetic energy of each of the particles and the potential energy stored in each of the three springs, and sum the results.) Notice that the kinetic and potential energy terms that have 12 as a coefficient depend on C1 and 2 but not on C2 or 2. Why is such a result to be expected?arrow_forwardIf a car has a suspension system with a force constant of 5.00104 N/m , how much energy must the car’s shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m?arrow_forward
- Consider the damped oscillator illustrated in Figure 12.16a. The mass of the object is 375 g, the spring constant is 100 N/m, and b = 0.100 N s/m. (a) Over what time interval does the amplitude drop to half its initial value? (b) What If? Over what time interval does the mechanical energy drop to half its initial value? (c) Show that, in general, the fractional rate at which the amplitude decreases in a damped harmonic oscillator is one-half the fractional rate at which the mechanical energy decreases.arrow_forwardConsider the system shown in Figure P16.68 as viewed from above. A block of mass m rests on a frictionless, horizontal surface and is attached to two elastic cords, each of length L. At the equilibrium configuration, shown by the dashed line, the cords both have tension FT. The mass is displaced a small amount as shown in the figure and released. Show that the net force on the mass is similar to the spring-restoring force and find the angular frequency of oscillation, assuming the mass behaves as a simple harmonic oscillator. You can assume the displacement is small enough to produce negligible change in the tension and length of the cords. FIGURE P16.68arrow_forwardConsider a graphical representation (Fig. 12.3) of simple harmonic motion as described mathematically in Equation 12.6. When the particle is at point on the graph, what can you say about its position and velocity? (a) The position and velocity are both positive. (b) The position and velocity are both negative. (c) The position is positive, and the velocity is zero. (d) The position is negative, and the velocity is zero. (e) The position is positive, and the velocity is negative. (f) The position is negative, and the velocity is positive. Figure 12.3 (Quick Quiz 12.2) An xt graph for a particle undergoing simple harmonic motion. At a particular time, the particles position is indicated by in the graph.arrow_forward
- An object of mass m1 = 9.00 kg is in equilibrium when connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in Figure P12.67a. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m (see Fig. P12.67b). The system is then released, and both objects start moving to the right on the frictionless surface. (a) When m1 reaches the equilibrium point, m2 loses contact with m1 (see Fig. P12.67c) and moves to the right with speed v. Determine the value of v. (b) How far apart are the objects when the spring is fully stretched for the first time (the distance D in Fig. P12.67d)? Figure P12.67arrow_forwardA spring 1.50 m long with force constant 475 N/m is hung from the ceiling of an elevator, and a block of mass 10.0 kg is attached to the bottom of the spring. (a) By how much is the spring stretched when the block is slowly lowered to its equilibrium point? (b) If the elevator subsequently accelerates upward at 2.00 m/s2, what is the position of the block, taking the equilibrium position found in part (a) as y = 0 and upwards as the positive y-direction. (c) If the elevator cable snaps during the acceleration, describe the subsequent motion of the block relative to the freely falling elevator. What is the amplitude of its motion?arrow_forwardA spherical bob of mass m and radius R is suspended from a fixed point by a rigid rod of negligible mass whose length from the point of support to the center of the bob is L (Fig. P16.75). Find the period of small oscillation. N The frequency of a physical pendulum comprising a nonuniform rod of mass 1.25 kg pivoted at one end is observed to be 0.667 Hz. The center of mass of the rod is 40.0 cm below the pivot point. What is the rotational inertia of the pendulum around its pivot point?arrow_forward
- A 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_forwardSuppose a diving board with no one on it bounces up and down in a SHM with a frequency of 4.00 Hz. The board has an effective mass of 10.0 kg. What is the frequency of the SHM of a 75.0-kg diver on the board?arrow_forwardA 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.arrow_forward
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