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The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The maximum speed of junction P is `V_0`. Assume springs are massless. Select the correct option/s A. Amplitude of potential energy is springs `S_(1)` and `S_(2)` is respectivley `1/9 kA^(2)` and `kA^(2)`B. Amplitude of potential energy is `S_(1)` and `S_(2)` is respectively `2/9 kA^(2)` and `1/9 kA^(2)`C. `V_(0)=Asqrt((8k)/(27 M))`D. `V_(0)=Asqrt((8k)/(81M))` |
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Answer» Correct Answer - BC `(2k)/3 A =kS_(p)` `Sp_(1)=2/3 A` `Sp_(2)=A/3` `U_("1 max") =1/2 k xx(4A^(2))/9=(2kA^(2))/9` `U_("2 max")=1/2 2 k xx(A^(2))/9=(kA^(2))/9` `V_(0)=(2A)/3xxsqrt((2k)/(3m))` |
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