Question

A simple harmonic oscillator consists of a block of mass 4.30 kg attached to a spring of spring constant 440 N/m. When t = 1.90 s, the position and velocity of the block are x = 0.179 m and v = 4.100 m/s. What is the amplitude of the oscillations? What were the position and velocity of the block at t = 0 s?

Answer 1

a) The displacement of the object in SHM is given as follows, x= x cos cos(ot+) (1) Here, x is the displacement, xm is the maximum amplitude, w is the angular frequency, is phase constant and t is the time. First calculate the angular frequency as follows, k = m2 Here, k is spring constant and m is the mass of block. Substitute 440 N/m for k, and 4.3 kg for m in above equation. Ik = m 440 4.3 = 10.1156 rad/s Now, find out the velocity of an object in SHM as follows, dx V=- dt d(x cos(@t+)) dt =-x.sin (@t+0) V=-x osin (ot +) (2)

From (1) and (2) v_-x@sin(wt+) xcos (ot+) =- tan(at +) X Now, at t = 1.90 s 4.100 =-(10.1156) tan (10.1156)(1.90)+0) 0.179 22.90=-(10.1156) tan (19.22+0) -2.26 = tan (19.22+0) (19.22+6)=-tan- (2.26) =-1.154 radian 0 = -20.374 radian Substitute the values in equation (1) x = x, cos(wt+) 0.179= x, cos((10.1156)(1.90) – 20.374 ) 0.179= x cos(19.21964 – 20.374) 0.179= x cos(-1.154 rad) = x(0.4045) 0.179 x = 0.4045 = 0.442 m Thus, the amplitude of oscillator is 0.442 m.

b) At t=0 the values are substituted in equation (1) Substitute 0.442 m for Xm and -20.374 rad for o and 0 s fort in below equation. x = x cos(@t+) =(0.442) cos(0-20.374 radian) = (0.442)(0.0463) = 0.0205 m c) The velocity of the block at t=0 s is, v=-x@sin (ot+0) =-(0.442) (10.1156) sin(-20.374 rad) = (4.47)(0.9989) = 4.466 m/s Therefore, the velocity of the block is 4.47 m/s.