The kinetic energy of a simple harmonic oscillator is oscillating with angular frequency of 176 rad/s$176 \text{ rad/s}$. The frequency of this simple harmonic oscillator is ________ Hz. [take π = (22)/(7)]$[\text{take } \pi = \frac{22}{7}]$

Related Formula

Core Logic

Squaring both sides of the simple pendulum equation:

Inverting T²$T^2$ to express 1/T²$1/T^2$ as a function of length L$L$:

This demonstrates that (1)/(T²) ∝ (1)/(L)$\frac{1}{T^2} \propto \frac{1}{L}$. The relationship between 1/T²$1/T^2$ and L$L$ represents a rectangular hyperbola curve decreasing asymptotically.

Step 1: Final Conclusion

Plot (2) correctly depicts the rectangular hyperbola curve for (1)/(T²)$\frac{1}{T^2}$ versus L$L$.

Pattern Recognition

Sees: Graph of 1/T²$1/T^2$ vs L$L$. Formula check: T ∝ √(L) T² ∝ L 1/T² ∝ 1/L$T \propto \sqrt{L} \implies T^2 \propto L \implies 1/T^2 \propto 1/L$, which is inversely proportional (hyperbolic curve).

Chapter Mix

Class 11 Physics: Simple Harmonic Motion

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Core Logic

For a simple pendulum, the period is proportional to the square root of the length. If the frequency is doubled, the period is halved, which requires the length to become one-fourth of its original value.

Step 1: Find Old and New Period

Initial time period T₁ = (10)/(20) = 0.5 s$T_{1} = \frac{10}{20} = 0.5 \text{ s}$ New time period T₂ = (10)/(40) = 0.25 s$T_{2} = \frac{10}{40} = 0.25 \text{ s}$ So, T₂ = T₁2$T_{2} = \frac{T_{1}}{2}$

Step 2: Calculate New Length

Pattern Recognition

Sees: "Oscillations doubled in same time" → Frequency is doubled → Time period is halved. Since T ∝ √(l)$T \propto \sqrt{l}$, l$l$ must be reduced to (1)/(4)$\frac{1}{4}$ of the original.

Chapter Mix

Class 11 Physics: Oscillations

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Core Logic

At equilibrium, the buoyant force balances the weight: ρ Ahg = Mg$\rho Ahg = Mg$ After displacing the block downward by a small distance x$x$, the net restoring force is:

Since ρ Ahg = Mg$\rho Ahg = Mg$, this simplifies to: Ma = -ρ Axg$Ma = -\rho Axg$

Step 1: Compare with Standard SHM Equation

Comparing with a = -ω² x$a = -\omega^2 x$, we get:

The time period T$T$ is:

Pattern Recognition

For a floating body of uniform cross-section A$A$, the restoring force constant is simply k = ρ A g$k = \rho A g$. The time period is immediately T = 2π √(m/k)$T = 2\pi \sqrt{m/k}$.

Chapter Mix

Class 11 Physics: Oscillations Class 11 Physics: Mechanical Properties of Fluids

Related Formula

PEmax$PE_{\text{max}}$ occurs when x = ± A$x = \pm A$.

Core Logic

Potential energy is maximum at the extreme positions. From x(t) = A (ω t)$x(t) = A \sin(\omega t)$, the particle starts at the mean position (x=0$x=0$ at t=0$t=0$) and reaches the extreme position (x=A$x=A$) for the first time at t = T/4$t = T/4$.

Step 1: Finding beta

Given time for maximum PE is t = (T)/(4)$t = \frac{T}{4}$. We are given t = (T)/(2β)$t = \frac{T}{2\beta}$. Therefore, (T)/(4) = (T)/(2β) ⇒ 2β = 4 ⇒ β = 2$\frac{T}{4} = \frac{T}{2\beta} \Rightarrow 2\beta = 4 \Rightarrow \beta = 2$.

Pattern Recognition

Sine function SHM starts at mean, hits extreme at T/4$T/4$, back to mean at T/2$T/2$, negative extreme at 3T/4$3T/4$, back to mean at T$T$. PE is max at T/4$T/4$ and 3T/4$3T/4$.

Chapter Mix

Class 11 Physics: Oscillations