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The simple harmonic oscillation of a particle are according to the equation x=5os(2pit+(pi)/4) metre. Find the (i) displacement (ii) velcoity and (iii) acceleration at t=0. |
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> :Comparing with the general equation <br/> `x=A <a href="https://interviewquestions.tuteehub.com/tag/cos-935872" style="font-weight:bold;" target="_blank" title="Click to know more about COS">COS</a> (omegat +phi_(0))` <br/> Amplitude `A=5m` and `omega=2pi` <br/> (i) Displacement `x=5 cos (2pit+(pi)/4)` metre <br/>when `t=0, x=x_(0)=5"cos"(pi)/4=5xx1/(sqrt(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))=3.54m` <br/> (ii) velocity `v=omega sqrt(A^(2)-x^(2))` <br/>when `t=0, v=v_(0)=omegasqrt(A^(2)-x_(0)^(2))` <br/> `=2pisqrt(5^(2)-(5/(sqrt(2)))^(2))=2pisqrt(25-25/2)` <br/> `=2pisqrt(25/2)=22.2ms^(-1)` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) Acceleration `a=-omega^(2)x` <br/> when `t=0=a=a_(0)=-omega^(2)-omega^(2)x_(0)`<br/> `=-(2pi)^(2)3.54=-4pi^(2)(3.54)^(2)` <br/> `:.a_(0)=139.42ms^(-2)`</body></html> | |