Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R) : Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements, choose the most appropriate answer from the options given below:

### Related Formula For a simple harmonic oscillator starting from the mean position at t = 0$t = 0$: x(t) = A sin(omega t)$$x(t) = A \sin(\omega t)$$ where omega = frac2piT$\omega = \frac{2\pi}{T}$ is the angular frequency. ### Core Logic Given: * Time period, T = 6pitext s implies omega = frac2pi6pi = frac13text rad/s$T = 6\pi\text{ s} \implies \omega = \frac{2\pi}{6\pi} = \frac{1}{3}\text{ rad/s}$ We need to find the time taken to travel from x = A$x = A$ to x = fracsqrt32A$x = \frac{\sqrt{3}}{2}A$. Let's calculate the times from the mean position (x=0$x=0$) to both positions: 1. Time t_1$t_1$ to reach x = A$x = A$: A = A sin(omega t_1) implies sin(omega t_1) = 1 implies omega t_1 = fracpi2$$A = A \sin(\omega t_1) \implies \sin(\omega t_1) = 1 \implies \omega t_1 = \frac{\pi}{2}$$ 2. Time t_2$t_2$ to reach x = fracsqrt32A$x = \frac{\sqrt{3}}{2}A$: fracsqrt32A = A sin(omega t_2) implies sin(omega t_2) = fracsqrt32 implies omega t_2 = fracpi3$$\frac{\sqrt{3}}{2}A = A \sin(\omega t_2) \implies \sin(\omega t_2) = \frac{\sqrt{3}}{2} \implies \omega t_2 = \frac{\pi}{3}$$ ### Step 1: Calculate the Time Difference The time required to travel between these two points is: Delta t = t_1 - t_2 = fracpi/2omega - fracpi/3omega = fracpi6omega$$\Delta t = t_1 - t_2 = \frac{\pi/2}{\omega} - \frac{\pi/3}{\omega} = \frac{\pi}{6\omega}$$ Substitute omega = frac13$\omega = \frac{1}{3}$: Delta t = fracpi6 times (1/3) = fracpi2text seconds$$\Delta t = \frac{\pi}{6 \times (1/3)} = \frac{\pi}{2}\text{ seconds}$$ Comparing this to fracpixtext s$\frac{\pi}{x}\text{ s}$, we find: x = 2$x = 2$

### Pattern Recognition Alternatively, using a phasor diagram, the angle swept in moving from A$A$ to fracsqrt32A$\frac{\sqrt{3}}{2}A$ is phi = fracpi6$\phi = \frac{\pi}{6}$. Therefore, Delta t = fracphiomega = fracpi/61/3 = fracpi2$\Delta t = \frac{\phi}{\omega} = \frac{\pi/6}{1/3} = \frac{\pi}{2}$ s. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula v_textmax = Aomega$$v_{\text{max}} = A\omega$$ v = omega sqrtA^2 - x^2$$v = \omega \sqrt{A^2 - x^2}$$ ### Core Logic From the mean position maximum criteria (v_textmax = 10text cm/s$v_{\text{max}} = 10\text{ cm/s}$, A = 4text cm$A = 4\text{ cm}$): 10 = 4omega implies omega = frac104 = frac52text rad/s$$10 = 4\omega \implies \omega = \frac{10}{4} = \frac{5}{2}\text{ rad/s}$$ ### Step 1: Substitute parameters into speed equation Set the target speed v = 5text cm/s$v = 5\text{ cm/s}$: 5 = frac52 sqrt4^2 - x^2$$5 = \frac{5}{2} \sqrt{4^2 - x^2}$$ 2 = sqrt16 - x^2$$2 = \sqrt{16 - x^2}$$ Squaring both sides: 4 = 16 - x^2 implies x^2 = 12 implies x = sqrt12text cm$$4 = 16 - x^2 \implies x^2 = 12 \implies x = \sqrt{12}\text{ cm}$$ ### Step 2: Compare to target format Comparing sqrt12$\sqrt{12}$ to sqrtalpha$\sqrt{\alpha}$ directly shows: alpha = 12$\alpha = 12$ ### Pattern Recognition Halving the peak harmonic velocity maps spatial points directly to fracsqrt32A$\frac{\sqrt{3}}{2}A$ values through standard trigonometric projection balances. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula Total energy (E$E$) in simple harmonic motion is: E = frac12 k A^2$$E = \frac{1}{2} k A^2$$ Potential energy (U$U$) at displacement y$y$ is: U = frac12 k y^2$$U = \frac{1}{2} k y^2$$ Kinetic energy (KE$KE$) is the remaining energy: KE = E - U$KE = E - U$ ### Core Logic Given the displacement is one-third of the amplitude: y = fracA3$$y = \frac{A}{3}$$ Substituting this displacement into the potential energy expression: U = frac12 k left(fracA3right)^2 = frac19 left( frac12 k A^2 right) = fracE9$$U = \frac{1}{2} k \left(\frac{A}{3}\right)^2 = \frac{1}{9} \left( \frac{1}{2} k A^2 \right) = \frac{E}{9}$$ ### Step 1: Calculate Kinetic Energy The kinetic energy is: $ ### Step 1: Calculate Kinetic Energy The kinetic energy is: $KE = E - fracE9 = frac8E9$KE = E - \frac{E}{9} = \frac{8E}{9}$ ### Step 2: Find the Energy Ratio The ratio of total energy to kinetic energy is: $ ### Step 2: Find the Energy Ratio The ratio of total energy to kinetic energy is: $fracEKE = fracEfrac8E9 = frac98$\frac{E}{KE} = \frac{E}{\frac{8E}{9}} = \frac{9}{8}$ Comparing this with the given target form $ Comparing this with the given target form $\frac{x}{8}: $: $x = 9 Therefore, the value of $ Therefore, the value of $x is $ is $9. ### Pattern Recognition Since $. ### Pattern Recognition Since $U \propto y^2, if displacement scales by a fraction $, if displacement scales by a fraction $\frac{1}{n}, potential energy scales instantly by $, potential energy scales instantly by $\frac{1}{n^2}. The remaining kinetic energy is naturally $. The remaining kinetic energy is naturally $1 - \frac{1}{n^2}, giving an immediate shortcut for the total-to-kinetic ratio: $, giving an immediate shortcut for the total-to-kinetic ratio: $\frac{n^2}{n^2 - 1}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations

### Related Formula g' = fracGM(R+h)^2 = gleft(fracRR+hright)^2$$g' = \frac{GM}{(R+h)^2} = g\left(\frac{R}{R+h}\right)^2$$ T = 2pi sqrtfraclg'$$T = 2\pi \sqrt{\frac{l}{g'}}$$ ### Core Logic The distance of the pendulum from the earth's surface is h = R$h = R$. The acceleration due to gravity at this height g'$g'$ is: g' = gleft(fracRR+Rright)^2 = gleft(frac12right)^2 = fracg4$$g' = g\left(\frac{R}{R+R}\right)^2 = g\left(\frac{1}{2}\right)^2 = \frac{g}{4}$$ ### Step 1: Calculate Time Period The time period of the simple pendulum is: T = 2pi sqrtfraclg'$$T = 2\pi \sqrt{\frac{l}{g'}}$$ Substitute l = 4 mathrm~m$l = 4 \mathrm{~m}$ and g' = fracg4$g' = \frac{g}{4}$: T = 2pi sqrtfrac4g/4 = 2pi sqrtfrac16g$$T = 2\pi \sqrt{\frac{4}{g/4}} = 2\pi \sqrt{\frac{16}{g}}$$ Given g = pi^2$g = \pi^2$: T = 2pi sqrtfrac16pi^2 = 2pi left(frac4piright) = 8 mathrm~s$$T = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \left(\frac{4}{\pi}\right) = 8 \mathrm{~s}$$ ### Pattern Recognition At h=R$h=R$, g$g$ drops to g/4$g/4$. Thus, the time period doubles compared to its surface value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Oscillations Class 11 Physics: Gravitation