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Question Number 87752 by Rio Michael last updated on 06/Apr/20

A particle exhibits simple hamornic motion such that    (d^2 x/dt^2 ) + 4x = 0   Calculate the period of the ocsillation

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{exhibits}\:\mathrm{simple}\:\mathrm{hamornic}\:\mathrm{motion}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\:+\:\mathrm{4}{x}\:=\:\mathrm{0} \\ $$$$\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ocsillation}\: \\ $$

Answered by TANMAY PANACEA. last updated on 06/Apr/20

(d^2 x/dt^2 )+w^2 x=0  w^2 =4  ((2π)/T)=2→T=πsecond

$$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+{w}^{\mathrm{2}} {x}=\mathrm{0} \\ $$$${w}^{\mathrm{2}} =\mathrm{4} \\ $$$$\frac{\mathrm{2}\pi}{{T}}=\mathrm{2}\rightarrow{T}=\pi{second} \\ $$

Commented by Rio Michael last updated on 06/Apr/20

thanks sir,please attempt the question above

$$\mathrm{thanks}\:\mathrm{sir},\mathrm{please}\:\mathrm{attempt}\:\mathrm{the}\:\mathrm{question}\:\mathrm{above} \\ $$

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