A 150 gram mass attached to a spring with a force constant of45.0 N/m vibrates o

Question

A 150 gram mass attached to a spring with a force constant of45.0 N/m vibrates on a horizontal, frictionless surface in simpleharmonic motion with an amplitude of 5.00cm. a. find the period of the oscillation b. find the total energy of the system c. find the speed of the mass when it is at positionx=2.00cm d.find the maximum speed and acceleration of the mass e.if at t=0 x=0 and if the mass has a positive velocity at t=0write the equation describing the position x of the mass as afunction of time, and find the time it takes for the mass to firstarrive at the position x=2.00cm A 150 gram mass attached to a spring with a force constant of45.0 N/m vibrates on a horizontal, frictionless surface in simpleharmonic motion with an amplitude of 5.00cm. a. find the period of the oscillation b. find the total energy of the system c. find the speed of the mass when it is at positionx=2.00cm d.find the maximum speed and acceleration of the mass e.if at t=0 x=0 and if the mass has a positive velocity at t=0write the equation describing the position x of the mass as afunction of time, and find the time it takes for the mass to firstarrive at the position x=2.00cm

Explanation / Answer

    Mass, M = 150 g                   = 0.150 kg     Force constant, K = 45 N/m     Amplitude, A = 5 cm                         = 0.05 m (a)     Force constant, K = M 2     Angular frequency, = ( K / M)                                     = ( 45 / 0.150 )                                     = 17.32 rad/s     Time-period, T = 2 /                            = 2 / 17.32                            = 0.363 s (b)     Total energy, E = ( 1/2 ) K A 2                            = 0.5 * 45 * 0.05 2                            = 0.0563 J (c)     Position, x = 2 cm                     = 0.02 m     Speed, V = ( A 2 - x2 )                   = 17.32 * [ ( 0.05 2 - 0.02 2 ) ]                   = 0.794 m/s (d)     Maximum speed, Vmax = A                                         = 17.32 * 0.05                                         = 0.866 m/s     Maximum acceleration, a max = 2 A                                                  = 17.32 2 * 0.05                                                  = 15 m/s^2 (e)     X = A cos t     X = 0.05 cos 17.32 t     When X = 0.02 m     0.02 = 0.05 cos 17.32 t     0.4 = cos 17.32 t     17.32 t = 1.159     Time, t = 0.067 s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.