question_answer 10)

In Q.9 above, let us take position of mass when the spring is unstretched as \[x=0\], and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the osculating mass if at the moment we start the stop watch \[(t=0)\], the mass is (a) at the mean position (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for S.H.M. differ from each other, in frequency, in amplitude or the initial phase?

Answer:

Here, \[2\cdot 0\] \[0\cdot 02\] (a) As time is noted from the mean position, hence using \[v=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{k}{m}}=\frac{1}{2\times 3\cdot 14}\sqrt{\frac{1200}{3}}=3\cdot 2{{s}^{-1}}\] (b) At maximum stretched position, the body is at the extreme right position, with an initial phase of \[\text{A=}{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{y=}\frac{\text{k}}{\text{m}}\text{y}\] rad. Then, \[\therefore \] (c) At maximum compressed position, the body is at the extreme left position, with an initial phase of\[{{\text{A}}_{\text{max}}}\text{=}\frac{\text{ka}}{\text{m}}=\frac{1200\times 0\cdot 02}{3}=8m{{s}^{-2}}\] \[{{\text{V}}_{\text{max}}}\text{=a }\!\!\omega\!\!\text{ =a}\sqrt{\frac{\text{k}}{\text{m}}}\text{=}0\cdot 02\times \sqrt{\frac{1200}{3}}=0\cdot 4\text{m}{{\text{s}}^{\text{-1}}}\]. The functions neither differ in amplitude nor in frequency. They differ in initial phase.