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Oscillations
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A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
Set $\omega\sqrt{A^2 - x^2} = \omega^2x$ to find ω.
$v = \omega\sqrt{A^2 - x^2}$, $a = x\omega^2$ $v = a \Rightarrow \omega\sqrt{A^2 - x^2} = x\omega^2$ $\sqrt{(3)^2 - (2)^2} = 2\left(\frac{2\pi}{T}\right)$ $\sqrt{5} = \frac{4\pi}{T}$ $T = \frac{4\pi}{\sqrt{5}}$ -
A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k''. Then k' : k'' is
Pieces have constants 6k, 3k, 2k; series gives k, parallel gives 11k.
Spring constant $\propto \frac{1}{length}$, i.e. $k \propto \frac{1}{l}$ $k_1 = 6k$, $k_2 = 3k$, $k_3 = 2k$ In series: $\frac{1}{k'} = \frac{1}{6k} + \frac{1}{3k} + \frac{1}{2k} = \frac{6}{6k} \Rightarrow k' = k$ In parallel: $k'' = 6k + 3k + 2k = 11k$ $\frac{k'}{k''} = \frac{1}{11}$, i.e. $k' : k'' = 1 : 11$