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 $k _{\text{eff }}=n^{2} k$.