Question

Review Constants Let's begin with a straightforward example of simple harmonic motion (SHM). A spring is mounted horizontally on an air track as in (Figure 1), with the left end held stationary. We attach a spring balance to the free end of the spring, pull toward the right, and measure the elongation. We determine that the stretching force is proportional to the displacement and that a force of 60 N causes an elongation of 0.030 m. We remove the spring balance and attach a 0.50 kg object to the end, pull it a distance of 0.040 m, release it, and watch it oscilate in SHM as in (Figure 2). Find the following quantities a. The force constant of the spring b. The maximum and minimum velocities attained by the vibrating object I SOLUTION SET UP AND SOLVE For this glider, the amplitude is A=0.040 m Part (a): When x = 0.030 m, the force the spring exerts on the object is F. =-6.0 N. From F, =-ker, the force constant kis F. -6.0 N ke 200 N/m 0.030 m Part (b): From Umut = k/m, the maximum velocity occurs at the equilibrium position, r=0. Te 200 N/m Umax = A 0.50 kg (0.040 m) = 0.80 m/s The minimum (most negative) velocity also occurs at z = 0 and is the negative of the maximum value. The minimum magnitude of velocity occurring at = ŁA, is zero Part(): The acceleration for any aris given by T2 Figure (1 of 2 SON m 0 0.030 The maximum (most positive) acceleration occurs at the most negative value of that is therefore, 200 N/m = A) 0.50 kg (-0.040 m) = 16.0 m/s? The minimum (most negative) acceleration is - 160 m/s occurring at +4 +0.010 m Part (d): The velocity at any position is given by

Answer 1

(A) A = 0.053 m í x = 0.016 m m=0.50 kg K = 200 N/m 2 KE = = a 늦m m ( JAP_x? 27? KE = < K (A²-x²) Ź (200) [10.053)– (0.016)?] 100[0.00 2809 -0.000256] KE - KE = KE = 0.2553 J