question_answer

If \[x=a\,\sin \,mt-b\,\cos \,mt\] and \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=\mu x,\] then find the value of \[\mu .\]

Answer:

We have, \[x=a\,\sin \,mt-b\,\cos \,mt\] On differentiating both sides w.r.t. ?t?, we get \[\frac{dx}{dt}=a\,\cos \,mt\,(m)-b\,(-\sin \,mt)(m)\] \[=am\,\cos \,mt+bm\,\sin \,mt\] Again differentiating w.r.t. ?t?, we get \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=am(-sin\,mt)(m)+bm(cos\,mt)(m)\] \[=-\,a{{m}^{2}}\sin \,mt+b{{m}^{2}}\cos \,mt\] \[=-\,{{m}^{2}}(a\,sin\,mt-b\,cos\,mt)=-{{m}^{2}}x\] According to the question, \[\frac{{{d}^{2}}x}{dt}=\mu x\]          \[\Rightarrow \]   \[-{{m}^{2}}x=\mu x\] \[\therefore \]      \[\mu =-\,{{m}^{2}}\]