A body executes simple harmonic motion under action of a force F₁ with a time period (4/5), sec .

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13.Oscillations

hard

A body executes simple harmonic motion under the action of a force $F_1$ with a time period $(4/5)\, sec$. If the force is changed to $F_2$ it executes $SHM$ with time period $(3/5)\, sec$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body, its time period (in $seconds$ ) is

$12/25$

$24/25$

$35/24$

$25/12$

Solution

$\mathrm{F}_{1}=-\mathrm{k}_{1} \mathrm{x}, \quad \mathrm{F}_{2}=-\mathrm{k}_{2} \mathrm{x}$

$\mathrm{f}_{1}=-\left(\frac{\mathrm{k}_{1}}{\mathrm{m}}\right) \mathrm{x} ; \quad \mathrm{f}_{2}=-\left(\frac{\mathrm{k}_{2}}{\mathrm{m}}\right) \mathrm{x}$

or $\quad \mathrm{f}_{1}=-\omega_{1}^{2} \mathrm{x}$ $\mathrm{f}_{2}=-\mathrm{w}_{2}^{2} \mathrm{x}$

Now, resultant force $\mathrm{F}=\mathrm{F}_{1}+\mathrm{F}_{2}=-\mathrm{k}_{1} \mathrm{x}-\mathrm{k}_{2} \mathrm{x}$

$\mathrm{Or}$ $-\mathrm{kx}=-\mathrm{k}_{1} \mathrm{x}-\mathrm{k}_{2} \mathrm{x}$

Or $\mathrm{k}=\mathrm{k}_{1}+\mathrm{k}_{2}$

Or $\mathrm{m} \omega^{2}=\mathrm{m} \omega_{1}^{2}+\mathrm{m} \omega_{2}^{2}$

Or $\omega^{2}=\omega_{1}^{2}+\omega_{2}^{2}$

Or $\left(\frac{2 \pi}{\mathrm{T}}\right)^{2}=\left(\frac{2 \pi}{\mathrm{T}_{1}}\right)^{2}+\left(\frac{2 \pi}{\mathrm{T}_{2}}\right)^{2}$

Or $\frac{1}{\mathrm{T}^{2}}=\frac{1}{\mathrm{T}_{1}^{2}}+\frac{1}{\mathrm{T}_{2}^{2}}$

$\mathrm{T}=\frac{\mathrm{T}_{1} \mathrm{T}_{2}}{\sqrt{\mathrm{T}_{1}^{2}+\mathrm{T}_{2}^{2}}}$

$=\frac{\frac{4}{5} \times \frac{3}{5}}{\sqrt{\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}}}=\frac{12}{25} \sec$

Standard 11

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Solution

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