Question

1.     a 10 g particle is undergoing simple harmonic motion with an amplitude of 2.0X10^-3 m and a maximum acceleration of magnitude 8.0X10^3 m/s^2. The phase constant is

Answer 1

given that

mass of the particle m = 10g = 10* 10^-3 kg

amplitude of particle   x_m = 2 * 10^-3 m

maximum accleration a( max) = 8 * 10^3 m/s^2 = 8000 m/s²

phase constant ? = -?/3

                           

from the maximum accleration

   a( max)= ?²x_m

   plugging above values ? = 2000 rad/s

a) from the newton second law F = ma

                                                  = m( -a( max) cos(?t+?)

                                                    = -80N cos ( 2000t -?/3)

b) T = 2?/?

            = 3.1 * 10^-3 s

spring constant k = ?² m = 40000 N/m

    from the conservation of energy

( 1/2) k x_m² = ( 1/2) mv_m²

      v_m = x_m ?k/m

            = 4.0 m/s

c) the total energy is ( 1/2) kx² = 1/2 * mv_m² = 0.080J