Simple Harmonic Motion

The time period of oscillation for a harmonic oscillator is given by.$ d 2 d d y=A \sin (\omega t)$$ $$ $

The amplitude of oscillation for a simple pendulum is represented by . The angular position of the pendulum as it swings is described by the equation:$ \theta_0 \omega \omega T L(t) I m l A L(t) \sin (\omega t) = \pm 1 l T $$ A $

Given,$ $ has a period also has a period To find the period of , we should determine the smallest positive such that .Since both and have a period of , we need to see how this affects . and since, has a period of , squaring it reduces the period by half to . retains the period since the cubic power does not change the period.The smallest common period of both and is the least common multiple (LCM) of and .Since is exactly twice , the LCM is .Therefore, the time period of the motion described by is:$ $

Given: Radius of the sphere: Half of the sphere is submerged in the pond.Solution:First, identify equilibrium conditions by equating the buoyant force and the sphere's weight. Next, consider the forces when the sphere is displaced: the additional buoyant force acts as the restoring force. Use this to set up the equation of motion for simple harmonic motion (SHM). Compare the resulting equation with the standard SHM form to find the angular frequency. Finally, derive the time period using the relationship between angular frequency and the time period of SHM.Buoyant Force and Equilibrium:The buoyant force when the sphere is at equilibrium (half-submerged) is equal to the weight of the sphere.$ F_b \Delta F_b=\rho_w \times A \times x \times g A=\pi R^2 x \rho_w=2 \rho F=m a a=y$$ m=\rho \times \frac{4}{3} \pi R^3 y=-\omega^2 x \omega $$ T T=\frac{2 \pi}{\omega} \omega $