Question

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes $ 3.5\times {{10}^{4}}K $ cm/s, will be :

• A. $ 5.8\times {{10}^{4}}K $
• B. $ \text{14 }\times \,\,\text{1}{{0}^{\text{4}}}\text{ K} $
• C. 1 cm
• D. 2 cm

Answer 1

Answer: (D)

At the mean position, the speed will be maximum $ =\frac{n\times \frac{1}{2}m{{\upsilon }^{2}}}{t} $ So, $ =\frac{360\times \frac{1}{2}\times 2\times {{10}^{-2}}\times {{(100)}^{2}}}{60} $ and $ g=\frac{Gm}{{{R}^{2}}} $ $ \left( \text{Here}:{{M}_{m}}=\frac{{{M}_{e}}}{9},{{R}_{m}}\frac{{{R}_{e}}}{2} \right) $ or $ \overrightarrow{\text{F}}=\left( \text{2\hat{i}}+\text{4\hat{j}} \right) $ or $ \overrightarrow{S}=\left( \text{3\hat{j}}+\text{5\hat{k}} \right)\text{m} $ or $ {{\upsilon }_{A}} $