There are two initial conditions associated with the simple harmonic motion differential equation. What is the meaning of this initial condition: a' (0) = B? O initial frequency O initial period O initial displacement O initial velocity
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- An object exhibiting simple harmonic motion has its position defined by the equation y = 5.00 sin (2t + pi/2) in meters. Find the acceleration of the particle at t = 2.81 s. (Hint: Find the second derivative of the position with respect to time and replace the value of t in the expression).What is the phase shift at the frequency of oscillation of each of the three sections? frequency = 100khzHelp me please
- Adapt the method of Example 6 to determine the amplitude, period, and phase shift for the given function. Graph the function over one period indicating the x- intercepts and the coordinates of the highest and lowest points on the graph. (a) y = 3 cos(7 - 2x) amplitude period phase shift y y 3- 3 3 3 5/7 Зя 4 4 4 4 4 2 o-3 y 4. 4 2 x-intercepts (х, у) - (smaller x-value) (х, у) - (larger x-value) highest points (x, y) = (smaller x-value) (x, y) = (larger x-value) lowest point (x, y) =The position of a particle is given by the expression x = 6.00 cos (4.00nt + 2π/5), where x is in meters and t is in seconds. (a) Determine the frequency. Hz (b) Determine period of the motion. S (c) Determine the amplitude of the motion. m (d) Determine the phase constant. rad (e) Determine the position of the particle at t = 0.350 s. m5. Review. A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration -0.320 m/s². Suppose it moves as a particle under constant acceleration for 4.50 s. Find (a) its position and Answer -2.34 m (b) its velocity at the end of this time interval. Next, assume it moves as a particle in simple harmonic motion for 4.50 s and x = 0 is its equilibrium position. Find Answer -1.30 m/s (c) its position and Answer -0.076 3 m (d) its velocity at the end of this time interval. Answer 0.315 m/s
- What is true about the acceleration of a simple harmonic oscillator? (Choose all correct answers; there may be more than one correct answer.) O The acceleration is a maximum when the object is instantaneously at rest. O The acceleration is a maximum when the displacement of the object is a maximum. O The acceleration is a maximum when the speed of the object is a maximum. O The acceleration is zero when the speed of the object is a maximum. O The acceleration is a maximum when the displacement of the object is zero.In simple harmonic motion, the displacement is maximum when the O velocity is zero O kinetic energy is maximum O displacement is zero O acceleration is zero O velocity is maximum O none of theseSimple harmonic motion can be described us- ing the equation y = Asin(kz – wt - 6). Consider the simple harmonic motion given by the figure. At position a= 0, we have +A What equation describes the motion? sin(-0)= - sin 0. 1. y = Asin (-wt +) 2. y = A cos (-wt - 3. y = Atan (-wt +5) 4. y = Acos (-wt+) 3m 5. y- Acos ( -wi+ 2 6. y - Alan 7. y = Atan -wt - 8. y = A sin (-wt - 9. y = Asin (-wt +) |