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(a) To double the total energy for a mass-spring system oscillating in
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Chapter 14 Solutions
University Physics with Modern Physics (14th Edition)
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- Show that angular frequency of a physical pendulum phy=mgrCM/I (Eq. 16.33) equals the angular frequency of a simple pendulum smp=g/, (Eq. 16.29) in the case of a particle at the end of a string of length .arrow_forwardWhich 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_forwardAn automobile with a mass of 1000 kg, including passengers, settles 1.0 cm closer to the road for every additional 100 kg of passengers. It is driven with a constant horizontal component of speed 20 km/h over a washboard road with sinusoidal bumps. The amplitude and wavelength of the sine curve are 5.0 cm and 20 cm, respectively. The distance between the front and back wheels is 2.4 m. Find the amplitude of oscillation of the automobile, assuming it moves vertically as an undamped driven harmonic oscillator. Neglect the mass of the wheels and springs and assume that the wheels are always in contact with the road.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_forwardDetermine the angular frequency of oscillation of a thin, uniform, vertical rod of mass m and length L pivoted at the point O and connected to two springs (Fig. P16.78). The combined spring constant of the springs is k(k = k1 + k2), and the masses of the springs are negligible. Use the small-angle approximation (sin ). FIGURE P16.78arrow_forwardCan this analogy of SHM to circular motion be carried out with an object oscillating on a spring vertically hung from the ceiling? Why or why not? If given the choice, would you prefer to use a sine function or a cosine function to model the motion?arrow_forward
- 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_forwardIf a simple pendulum oscillates with small amplitude and its length is doubled, what happens to the frequency of its motion? (a) It doubles. (b) It becomes 2 times as large. (c) It becomes half as large. (d) It becomes 1/2 times as large. (e) It remains the same.arrow_forwardA 500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. Calculate the maximum value of its (a) speed and (b) acceleration, (c) the speed and (d) the acceleration when the object is 6.00 cm from the equilibrium position, and (e) the time interval required for the object to move from x = 0 to x = 8.00 cm.arrow_forward
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