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A mass m is performing linear simple harmonic motion, then which of the following graph represents correctly the variation of acceleration a corresponding to linear velocity `upsilon` ?A. B. C. D. |
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Answer» Correct Answer - D In SHM, `y=rsin omegat` `upsilon=romega cos omegat and a=(dupsilon)/(dt)=-omega^(2)rsinomegat` `:. cos omegat =(upsilon)/(romega) and sin omegat =-(a)/(omega^(2)r)` Hence, `cos^(2)omegat + sin^(2)omegat=((upsilon)/(romega))^(2)+[-(a)/(omega^(2)r)]^(2)` or `1=(upsilon^(2))/(r^(2)omega^(2))+(a^(2))/(omega^(4)r^(2))` or ` upsilon^(2)=-(a^(2))/(omega^(2))+r^(2)omega^(2)` ...(i0 Comparing (i), with general equation of st. line, `y=mx+c` , we note theat, the equation (i) is a st. line betweent `upsilon^(2) and a^(2)` with negative slope `(1//omega^(2))`. Hence, optiond (d) is correct. |
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