4. (a) lua simple harmonic motion the magnitude of the force is given as F=-kx where is the force constant and is the displacement of the mass m (which is making the SHM) from its equilibrium position. (i) Find the differential equation for the above simple harmonic motion. (2 marks) (ii) Solve the above differential equation by taking its general solution as x=Acos(omega t+varphi ) and find the value of a in terms of k and m. (3 marks) (iii) What is the value of maximum displacement? (1 mark) (iv) Find the time period T for the SHM. (3 marks) (v) Find the potential and kinetic energies in the SHM and hence show that the total energy in a SHM is constant and is proportional to the square of the amplitude ot (5 marks)

(i) The differential equation for simple harmonic motion is derived from Newton's second law, \( F = ma \), where \( a \) is the acceleration. Given \( F = -kx \), we have:<br /><br />\[ ma = -kx \]<br /><br />Since acceleration \( a \) is the second derivative of displacement \( x \) with respect to time \( t \), i.e., \( a = \frac{d^2x}{dt^2} \), the equation becomes:<br /><br />\[ m\frac{d^2x}{dt^2} = -kx \]<br /><br />Rearranging gives the differential equation:<br /><br />\[ \frac{d^2x}{dt^2} + \frac{k}{m}x = 0 \]<br /><br />(ii) To solve this differential equation, assume the general solution \( x = A \cos(\omega t + \varphi) \). The second derivative is:<br /><br />\[ \frac{d^2x}{dt^2} = -A\omega^2 \cos(\omega t + \varphi) \]<br /><br />Substitute into the differential equation:<br /><br />\[ -A\omega^2 \cos(\omega t + \varphi) + \frac{k}{m}A \cos(\omega t + \varphi) = 0 \]<br /><br />This simplifies to:<br /><br />\[ A(-\omega^2 + \frac{k}{m}) \cos(\omega t + \varphi) = 0 \]<br /><br />For this to hold for all \( t \), we require:<br /><br />\[ \omega^2 = \frac{k}{m} \]<br /><br />Thus, \( \omega = \sqrt{\frac{k}{m}} \).<br /><br />(iii) The maximum displacement in SHM is the amplitude \( A \).<br /><br />(iv) The time period \( T \) of the SHM is related to the angular frequency \( \omega \) by:<br /><br />\[ T = \frac{2\pi}{\omega} \]<br /><br />Substituting \( \omega = \sqrt{\frac{k}{m}} \):<br /><br />\[ T = 2\pi \sqrt{\frac{m}{k}} \]<br /><br />(v) The potential energy \( U \) in SHM is given by:<br /><br />\[ U = \frac{1}{2}kx^2 \]<br /><br />The kinetic energy \( K \) is:<br /><br />\[ K = \frac{1}{2}mv^2 \]<br /><br />where \( v = \frac{dx}{dt} = -A\omega \sin(\omega t + \varphi) \), so:<br /><br />\[ K = \frac{1}{2}m(A\omega \sin(\omega t + \varphi))^2 = \frac{1}{2}mA^2\omega^2 \sin^2(\omega t + \varphi) \]<br /><br />Using \( \omega^2 = \frac{k}{m} \), we get:<br /><br />\[ K = \frac{1}{2}kA^2 \sin^2(\omega t + \varphi) \]<br /><br />The total energy \( E \) is the sum of potential and kinetic energies:<br /><br />\[ E = U + K = \frac{1}{2}kA^2 \cos^2(\omega t + \varphi) + \frac{1}{2}kA^2 \sin^2(\omega t + \varphi) \]<br /><br />\[ E = \frac{1}{2}kA^2 (\cos^2(\omega t + \varphi) + \sin^2(\omega t + \varphi)) \]<br /><br />Since \( \cos^2(\theta) + \sin^2(\theta) = 1 \), we have:<br /><br />\[ E = \frac{1}{2}kA^2 \]<br /><br />This shows that the total energy in SHM is constant and proportional to the square of the amplitude \( A \).