Oscillations - Result Question 38
43. A particle is executing a simple harmonic motion. Its maximum acceleration is a and maximum velocity is $b$.
[2015 RS]
Then its time period of vibration will be :
(a) $\frac{\alpha}{\beta}$
(b) $\frac{\beta^{2}}{\alpha}$
(c) $\frac{2 \pi \beta}{\alpha}$
(d) $\frac{\beta^{2}}{\alpha^{2}}$
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Answer:
Correct Answer: 43. (c)
Solution:
When they are connected in series
$\frac{1}{k^{\prime}}=\frac{1}{6 k}+\frac{1}{3 k}+\frac{1}{2 k}$
$\Rightarrow \frac{1}{k^{\prime}}=\frac{6}{6 k}$
$\therefore$ Force constant $k^{\prime}=k$
And when they are connected in parallel $k^{\prime \prime}=6 k+3 k+2 k$
$\Rightarrow k^{\prime \prime}=11 k$
Then the ratios
$\frac{k^{\prime}}{k^{\prime \prime}}=\frac{1}{11}$ i.e., $k^{\prime}: k^{\prime \prime}=1: 11$
If a spring of force constant $K$ is divided into $n$ equal parts then spring constant of each part will become $n k$ and if these $n$ parts connected in parallel then $k _{\text{eff }}=n^{2} k$.
