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)

2 cm

Answer 2

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 ${{\upsilon }_{A}}$