A spring of negligible mass stretches 3 cm from its relaxed lengthwhen a force o

Question

A spring of negligible mass stretches 3 cm from its relaxed lengthwhen a force of 8.0 N is applied. A0.4 kg particle rests on a frictionlesshorizontal surface and is attached to the free end of the spring.The particle is displaced from the origin to x = 5.00 cmand released from rest at t = 0. (a) What is the forceconstant of the spring?
1 N/m

(b) What are the angular frequency (), thefrequency, and the period of the motion? angular frequency 2 rad/s frequency 3 Hz period 4 s
(c) What is the total energy of the system?
5 J

(d) What is the amplitude of the motion?
6 cm

(e) What are the maximum velocity and the maximum acceleration ofthe particle? maximum velocity 7 m/s maximum acceleration 8 m/s2
(f) Determine the displacement, x, of the particle fromthe equilibrium position at t = 0.500 s.
9 cm

(g) Determine the velocity and acceleration of the particle whent = 0.500 s.
velocity 10 m/s acceleration 11 m/s2 angular frequency 2 rad/s frequency 3 Hz period 4 s

Explanation / Answer

  a.    Springconstant   k   =   forceapplied / extensionproduced   =   8.0 /0.03   =   266.67   N/m        b.   Angularfrequency      =   (k/m)   =   (266.67/0.40)   =   25.82   rad/s          Freuqncy   f   =   /2   =   25.82/ 2 *3.14   =   4.11   Hz          Period   T   =   1/f   =   1/4.11   =   0.243   s    c.   Totalenergy   E   =   (1/2)* m * 2 *A2,         A   =   Amplitude   =   0.05   m          E   =   0.5* 0.4 * 25.822 *0.052   =   0.333   J    d.   Amplitude   A   =   Maximumdisplacement      =   5.00cm      =   0.05   m    e.   Maximumvelocity   vmax   =         * A   =   25.82 *0.05   =   1.291   m/s          Maximumacceleration   max   =   -2 * A   =   -25.822 *1.291   =   -33.33   m/s2       -ve sign indicates thatthe acceleration is directed towards the origin.    f.   Equation of motionis   x   =   A * sin( * t + )             here      =   initialphase   =   900      asthe particle starts at one extreme          x   =   0.05* sin (25.82 *0.50   +   900)   =   0.05* cos 12.91   =   0.0471m   =   4.71   cm    g.   Velocity   v      =   * A * cos (t +)   =   25.82 * 0.05 * cos(25.82 *0.5   +   900)          v   =   0.435   m/s          Accleration      =   -2 * x   =   -25.822 *0.0471   =  - 31.40   m/s2       -ve sign indicates thatthe acceleration is directed towards the origin.    f.   Equation of motionis   x   =   A * sin( * t + )             here      =   initialphase   =   900      asthe particle starts at one extreme          x   =   0.05* sin (25.82 *0.50   +   900)   =   0.05* cos 12.91   =   0.0471m   =   4.71   cm    g.   Velocity   v      =   * A * cos (t +)   =   25.82 * 0.05 * cos(25.82 *0.5   +   900)          v   =   0.435   m/s          Accleration      =   -2 * x   =   -25.822 *0.0471   =  - 31.40   m/s2    

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