Question

35 I) The nearby graph shows position in cmr(t) vs. time in s. From the graph, estimate the (a) equilibrium position (b) amplitude (c) period of the oscillatory motion. For this system, determine the (d) cycle frequency (e) angular frequency (f) Write down an expression for this trajec- tory in terms of a sine function. 30 25 20 15 10 20 21 22 23 24 25 26 27 28 29 30 31

Answer 1

Given this graph is an example of a sine wave. We can simply say that it�s a S.H.S, simple harmonic motion. In S.H.M, the equilibrium position is where particle has no displacement. So simply at x = 0 i.e, horizontal axis will be the equilibrium position. Amplitude is given by the height of crest from the equilibrium point. So, Amplitude = 30 - 20 = 10cm time period = time taken to complete one cyclic then from graph. T = t(27 se)- t(21 sec) = 27 - 21 = 6 sec cyclic frequency = no.of cycle completed in one second. In 6 sec, cycle completed = 1 in 1 sec, cycle completed = 1/6 and we know f = 1/T = 1/6 \r\n angular frequency: (w) we know angular frequency is related to cyclic frequency by relation. W = 2 pi f = 2 pi 1/6 = pi/3 rad/sec the given graph is similar to sine function, so we can express the trajectory