Question

The period of small oscillations of a simple pendulum of length $\phi$, if its point of suspension O moves a with a constant acceleration $\alpha ={\alpha }_1\hat{i}-{\alpha }_2\hat{j}$ with respect to earth, is

• A. $T=2\pi \sqrt{\frac{\phi }{{\{{(g-{\alpha }_2)}^2+{\alpha }_1^2\}}^{1/2}}}$
• B. $T=2\pi \sqrt{\frac{\phi }{{\{{(g-{\alpha }_1)}^2+{\alpha }_2^2\}}^{1/2}}}$
• C. $T=2\pi \sqrt{\frac{\phi }{g}}$
• D. $T=2\pi \sqrt{\frac{\phi }{(g^2+{\alpha }_1^2{)}^{1/2}}}$

Answer 1

The correct option is A $T=2\pi \sqrt{\frac{\phi }{{\{{(g-{\alpha }_2)}^2+{\alpha }_1^2\}}^{1/2}}}$

Let us assume this is a simple pendulum.

As point 0 has acceleration

$a={\alpha }_1\hat{i}-{\alpha }_2\hat{j}$

So, pseudo force will act on Pendulum's bob .

$g_{\text{eff}}=-{\alpha }_1\hat{i}+({\alpha }_2-g)\hat{j}$

$\begin{array}{c}|geff|=\sqrt{{\alpha }_1^2+{(g-{\alpha }_2)}^2}\\ T=2\pi \sqrt{\frac{\phi }{geff}}\end{array}$

$=2\pi \sqrt{{\frac{\phi }{({\alpha }_1^2+{(g-{\alpha }_2)}^2}\}}^{1/2}}$