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The initial position, velocity, and acceleration of an object moving in
(b) Using A to represent the amplitude of the motion, show that
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Physics for Scientists and Engineers with Modern Physics
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- Which of the following statements is not true regarding a massspring system that moves with simple harmonic motion in the absence of friction? (a) The total energy of the system remains constant. (b) The energy of the system is continually transformed between kinetic and potential energy. (c) The total energy of the system is proportional to the square of the amplitude. (d) The potential energy stored in the system is greatest when the mass passes through the equilibrium position. (e) The velocity of the oscillating mass has its maximum value when the mass passes through the equilibrium position.arrow_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_forwardA block of unknown mass is attached to a spring with a spring constant of 6.50 N/m and undergoes simple harmonic motion with an amplitude of 10.0 cm. When the block is halfway between its equilibrium position and the end point, its speed is measured to be 30.0 cm/s. Calculate (a) the mass of the block, (b) the period of the motion, and (c) the maximum acceleration of the block.arrow_forward
- We do not need the analogy in Equation 16.30 to write expressions for the translational displacement of a pendulum bob along the circular arc s(t), translational speed v(t), and translational acceleration a(t). Show that they are given by s(t) = smax cos (smpt + ) v(t) = vmax sin (smpt + ) a(t) = amax cos(smpt + ) respectively, where smax = max with being the length of the pendulum, vmax = smax smp, and amax = smax smp2.arrow_forwardThe amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?arrow_forwardA 50.0-g object connected to a spring with a force constant of 35.0 N/m oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface. Find (a) the total energy of the system and (b) the speed of the object when its position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy when its position is 3.00 cm.arrow_forward
- The angular position of a pendulum is represented by the equation = 0.032 0 cos t, where is in radians and = 4.43 rad/s. Determine the period and length of the pendulum.arrow_forwardConsider the simplified single-piston engine in Figure CQ12.13. Assuming the wheel rotates with constant angular speed, explain why the piston rod oscillates in simple harmonic motion. Figure CQ12.13arrow_forwardIf the amplitude of a damped oscillator decreases to 1/e of its initial value after n periods, show that the frequency of the oscillator must be approximately [1 − (8π2n2)−1] times the frequency of the corresponding undamped oscillator.arrow_forward
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