Answer:
Here, r = 5 cm = 0.05 m ; T = 0.2 s;
\[\alpha \]
When displacement is y, then
acceleration,
\[\text{I=}\frac{\text{1}}{\text{2}}\text{m}{{\text{R}}^{\text{2}}}\text{=}\frac{1}{2}\times
10\times {{\left( 0\cdot 15 \right)}^{2}}kg{{m}^{2}}\]
velocity, \[\text{T=2
}\!\!\pi\!\!\text{ }\sqrt{\frac{\text{I}}{\text{ }\!\!\alpha\!\!\text{ }}}\]
Case (a) When y = 5 cm = \[\alpha =\frac{4{{\pi
}^{2}}l}{{{T}^{2}}}=4\times {{\left( \frac{22}{7} \right)}^{2}}\times
\frac{1}{2}\times \frac{10\times {{\left( 0\cdot 15 \right)}^{2}}}{{{\left(
1\cdot 5 \right)}^{2}}}\] m
\[=1\cdot 97Nm/rad.\]
Case (b) When y = 3 cm = \[0\cdot 05\] m
\[\cdot \]
Case (c) When\[y=0,\,A=-{{(10\pi
)}^{2}}\times 0=0\]
\[\text{A=-}{{\text{ }\!\!\omega\!\!\text{
}}^{\text{2}}}\text{y}\]
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