warehouse/content/physics/final.md

[!NOTE] A mass m attached to a spring with constant k oscillates on a frictionless surface. Derive an expression for the velocity v as a function of displacement x.

[!INFO]- Total mechanical energy in SHM is conserved:

E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2

Solving for v:

v = \pm \sqrt{\frac{k}{m}(A^2 - x^2)}

[!NOTE] A wave traveling along a rope is represented by y(x,t) = 0.02\cos(40x - 600t). Determine the amplitude, wavelength, frequency, and speed of the wave.

[!INFO]- The general wave form is y = A\cos(kx - \omega t):

• Amplitude A = 0.02\ \mathrm{m}
• Wave number k = 40 \Rightarrow \lambda = \frac{2\pi}{k} = \frac{2\pi}{40} = 0.157\ \mathrm{m}
• Angular frequency \omega = 600 \Rightarrow f = \frac{\omega}{2\pi} = \frac{600}{2\pi} \approx 95.5\ \mathrm{Hz}
• Wave speed v = f\lambda = 95.5 \times 0.157 \approx 15\ \mathrm{m/s}

[!NOTE] In Young's double slit experiment, fringes are formed on a screen 1.2 m away using light of wavelength 600\ \text{nm}. The slits are separated by 0.2\ \text{mm}. Calculate the distance between adjacent bright fringes.

[!INFO]- Fringe spacing is given by:

y = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1.2}{0.2 \times 10^{-3}} = 3.6\ \text{mm}

[!NOTE] A pendulum of length 0.5\ \mathrm{m} is displaced by a small angle. Determine its period and explain why amplitude does not affect the result.

[!INFO]- The period is:

T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{0.5}{9.8}} \approx 1.41\ \mathrm{s}

In the small-angle approximation (\theta < 10^\circ), motion is independent of amplitude.

[!NOTE] A charged particle q moves through a uniform electric field E. Derive the expression for the work done on the charge and its change in potential energy.

[!INFO]- Work done:

W = qEd

Change in potential energy:

\Delta U = -qEd

since electric potential energy decreases when the charge moves in the direction of the field.

[!NOTE] A capacitor of 10\ \mu\mathrm{F} is charged to 5\ \mathrm{V}. Calculate the energy stored in it.

[!INFO]-

E = \frac{1}{2}CV^2 = \frac{1}{2} \cdot 10 \times 10^{-6} \cdot 25 = 1.25 \times 10^{-4}\ \mathrm{J}

[!NOTE] A coil of wire rotates in a magnetic field. Use Faraday’s law to derive the expression for the induced emf.

[!INFO]- Faraday’s law:

\mathcal{E} = -\frac{d\Phi}{dt}

Magnetic flux \Phi = B A \cos(\omega t), so:

\mathcal{E} = B A \omega \sin(\omega t)

[!NOTE] A wire carries a current of 3\ \mathrm{A} through a magnetic field of 0.5\ \mathrm{T} perpendicular to its length. The wire is 0.4\ \mathrm{m} long. Find the magnetic force.

[!INFO]-

F = ILB\sin\theta = 3 \cdot 0.4 \cdot 0.5 \cdot 1 = 0.6\ \mathrm{N}

[!NOTE] Explain the effect of damping on the amplitude-frequency graph of a driven harmonic oscillator.

[!INFO]- Damping reduces the peak amplitude and shifts the resonant frequency slightly lower. Greater damping broadens the curve and lowers the quality factor Q.

[!NOTE] A mass-spring oscillator experiences light damping. Write the differential equation and general solution.

[!INFO]- Equation:

m\ddot{x} + b\dot{x} + kx = 0

Solution:

x(t) = A e^{-\gamma t} \cos(\omega' t + \phi)

where \gamma = \frac{b}{2m} and \omega' = \sqrt{\omega_0^2 - \gamma^2}.

[!NOTE] A 0.2\ \mathrm{kg} object experiences a force due to gravity from Earth at a distance of 6.4 \times 10^6\ \mathrm{m}. Calculate the force.

[!INFO]-

F = G\frac{Mm}{r^2} = 6.67 \times 10^{-11} \cdot \frac{5.97 \times 10^{24} \cdot 0.2}{(6.4 \times 10^6)^2} \approx 1.96\ \mathrm{N}