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Chapter 14 Solutions
University Physics with Modern Physics (14th Edition)
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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_forwardA mass is placed on a frictionless, horizontal table. A spring (k=100N/m) , which can be stretched or compressed, is placed on the table. A 5.00-kg mass is attached to one end of the spring, the other end is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x=4.0 cm and releases it from rest. The mass oscillates in SHM. (a) Determine the equations of motion. (b) Find the position, velocity, and acceleration of the mass at time t=3.00 s.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_forward
- A 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_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_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
- 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_forwardIn an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x=5.00cos(2t+6) where x is in centimeters and t is in seconds. At t = 0, find (a) the position of the piston, (b) its velocity, and (c) its acceleration. Find (d) the period and (e) the amplitude of the motion.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
- A nylon siring has mass 5.50 g and length L = 86.0 cm. The lower end is tied to the floor, and the upper end is tied to a small set of wheels through a slot in a track on which the wheels move (Fig. P18.76). The wheels have a mass that is negligible compared with that of the siring, and they roll without friction on the track so that the upper end of the string is essentially free. Figure P18.76 At equilibrium, the string is vertical and motionless. When it is carrying a small-amplilude wave, you may assume the string is always under uniform tension 1.30 N. (a) Find the speed of transverse waves on the siring, (b) The string's vibration possibilities are a set of standing-wave states, each with a node at the fixed bottom end and an antinode at the free top end. Find the node-antinode distances for each of the three simplest states, (c) Find the frequency of each of these states.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_forwardThe 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_forward
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