Chapter 15: Q94P (page 441)
What is the phase constant for SMH with given in Fig. if the position functionhas the formand?
Short Answer
The phase constant for SHM with given in the figure is .
Chapter 15: Q94P (page 441)
What is the phase constant for SMH with given in Fig. if the position functionhas the formand?
The phase constant for SHM with given in the figure is .
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Get started for freeFigure 15-25shows plots of the kinetic energy K versus position x for three harmonic oscillators that have the same mass. Rank the plots according to (a) the corresponding spring constant and (b) the corresponding period of the oscillator, greatest first.
A block hangs from a spring, extending it 16.0 cmfrom its unstretched position.
A particle executes linear SHM with frequency about the point . At, it has displacement and zero velocity. For the motion, determine the (a) period, (b) angular frequency, (c) amplitude, (d) displacement x(t), (e) velocity v(t), (f) maximum speed, (g) magnitude of the maximum acceleration, (h) displacement at , and (i) speed at.
A torsion pendulum consists of a metal disk with a wire running through its center and soldered in place. The wire is mounted vertically on clamps and pulled taut. 15-58a Figuregives the magnitude of the torque needed to rotate the disk about its center (and thus twist the wire) versus the rotation angle . The vertical axis scale is set by =.The disk is rotated to rad and then released. Figure 15-58bshows the resulting oscillation in terms of angular position versus time t. The horizontal axis scale is set by . (a) What is the rotational inertia of the disk about its center? (b) What is the maximum angular speedof the disk? (Caution: Do not confuse the (constant) angular frequency of the SHM with the (varying) angular speed of the rotating disk, even though they usually have the same symbol. Hint: The potential energy U of a torsion pendulum is equal to , analogous to for a spring.)
The center of oscillation of a physical pendulum has this interesting property: If an impulse (assumed horizontal and in the plane of oscillation) acts at the center of oscillation, no oscillations are felt at the point of support. Baseball players (and players of many other sports) know that unless the ball hits the bat at this point (called the “sweet spot” by athletes), the oscillations due to the impact will sting their hands. To prove this property, let the stick in Fig. simulate a baseball bat. Suppose that a horizontal force (due to impact with the ball) acts toward the right at P, the center of oscillation. The batter is assumed to hold the bat at O, the pivot point of the stick. (a) What acceleration does the point O undergo as a result of? (b) What angular acceleration is produced by about the center of mass of the stick? (c) As a result of the angular acceleration in (b), what linear acceleration does point O undergo? (d) Considering the magnitudes and directions of the accelerations in (a) and (c), convince yourself that P is indeed the “sweet spot.
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