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In Exercise 9, let us take theposition 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 oscillating mass if at the moment we start the stopwatch (t=0), the mass is at the mean position.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Amplitude from exercise : `14.9, A = 2 cm` <br/> Angular frequency `omega = sqrt((k)/(m)) = sqrt((1200)/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)) = sqrt(400) = 20 <a href="https://interviewquestions.tuteehub.com/tag/rad-1175618" style="font-weight:bold;" target="_blank" title="Click to know more about RAD">RAD</a> s^(-1)` <br/> Initial <a href="https://interviewquestions.tuteehub.com/tag/phase-22748" style="font-weight:bold;" target="_blank" title="Click to know more about PHASE">PHASE</a> at mean position `phi = <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>` <br/> Displacement of block at time t <br/> `x(t) = A sin(omega t+phi)` putting the abovevalues <br/> `= 2 sin(2 omega t +0)` <br/> ` =2 sin 20 t cm""[therefore omega = 20 rad s^(-1)]`</body></html>


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