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 maximum streatched position.

In Exercise 9, let us take theposition of mass when the spring is unst

1.	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 maximum streatched position.
Answer» <html><body><p></p>Solution :Amplitude from exercise : `14.9, A = 2 cm` <br/> <a href="https://interviewquestions.tuteehub.com/tag/angular-11524" style="font-weight:bold;" target="_blank" title="Click to know more about ANGULAR">ANGULAR</a> 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 rad s^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)` <br/> Phase in the maximum stretched position `phi =(pi)/(2)rad` <br/> Displacement of block at time t <br/> `x(t) =A sin (omega t+phi)` putting thevalue ` = <a href="https://interviewquestions.tuteehub.com/tag/2sin-301148" style="font-weight:bold;" target="_blank" title="Click to know more about 2SIN">2SIN</a> (2omega t+(pi)/(2))= 2 cos 20 t cm`.</body></html>