The energy density of an EM wave is proportional to the square of the electric f ield. Squaring a sinusoidal function doubles the frequency.
The energy density of the wave is given by:
\( \text{Energy density} = \frac{1}{2} \varepsilon_0 E_{\text{net}}^2 \)
Substitute \( E_{\text{net}} = E_0 \sin(\omega t - kx) \):
\[ \text{Energy density} = \frac{1}{2} \varepsilon_0 E_0^2 \sin^2(\omega t - kx) \]
Using the trigonometric identity \( \sin^2 x = \frac{1}{2}(1 - \cos 2x) \):
\[ \sin^2(\omega t - kx) = \frac{1}{2}(1 - \cos(2\omega t - 2kx)) \]
Substitute this into the energy density formula:
\[ \text{Energy density} = \frac{1}{2} \varepsilon_0 E_0^2 \cdot \frac{1}{2}(1 - \cos(2\omega t - 2kx)) \]
Simplify the expression:
\[ \text{Energy density} = \frac{1}{4} \varepsilon_0 E_0^2 (1 - \cos(2\omega t - 2kx)) \]
The energy density of the wave is:
\( \frac{1}{4} \varepsilon_0 E_0^2 (1 - \cos(2\omega t - 2kx)) \)
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: