Using a simple pendulum experiment g is determined by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T?

### Related Formula A = sqrtA_1^2 + A_2^2 + 2 A_1 A_2 cosphi$$A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos\phi}$$ a_max = omega^2 A$$a_{\max} = \omega^2 A$$ ### Core Logic Both SHMs share the same frequency omega = 5mathrm~rad/s$\omega = 5\mathrm{~rad/s}$ with phase difference phi = fracpi3$\phi = \frac{\pi}{3}$. Amplitudes: A_1 = sqrt7mathrm~cm$A_1 = \sqrt{7}\mathrm{~cm}$, A_2 = 2sqrt7mathrm~cm$A_2 = 2\sqrt{7}\mathrm{~cm}$. The combined amplitude is: A = sqrt(sqrt7)^2 + (2sqrt7)^2 + 2(sqrt7)(2sqrt7)cosleft(fracpi3right)$$A = \sqrt{(\sqrt{7})^2 + (2\sqrt{7})^2 + 2(\sqrt{7})(2\sqrt{7})\cos\left(\frac{\pi}{3}\right)}$$ A = sqrt7 + 28 + 2(7)(2)left(frac12right) = sqrt35 + 14 = sqrt49 = 7mathrm~cm$$A = \sqrt{7 + 28 + 2(7)(2)\left(\frac{1}{2}\right)} = \sqrt{35 + 14} = \sqrt{49} = 7\mathrm{~cm}$$ Converting to meters: A = 0.07mathrm~m$$A = 0.07\mathrm{~m}$$ The maximum acceleration is: a_max = omega^2 A = (5)^2 times 0.07 = 25 times 0.07 = 1.75mathrm~ms^-2$$a_{\max} = \omega^2 A = (5)^2 \times 0.07 = 25 \times 0.07 = 1.75\mathrm{~ms}^{-2}$$ Equating to x times 10^-2mathrm~ms^-2$x \times 10^{-2}\mathrm{~ms}^{-2}$: 1.75 = x times 10^-2 implies x = 175$$1.75 = x \times 10^{-2} \implies x = 175$$ ### Step 1: Final Conclusion The value of x$x$ is 175. ### Pattern Recognition When superposing two SHMs of identical frequency, use the standard phasor addition formula to obtain the combined amplitude A$A$. Then compute maximum velocity v_max = omega A$v_{\max} = \omega A$ or maximum acceleration a_max = omega^2 A$a_{\max} = \omega^2 A$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Simple Harmonic Motion