Question

A spring of spring constant 2000N/m is suspended vertically in earth's gravitaional field and a mass of 0.8 kg is hung from it. It is hanging in equilibrium .At t=0, a force of 72 N starts acting in vertically downward direction. The force ceases to act at $t = \frac{\pi}{2} sec$.If the amplitude of resulting SHM (in mm) after that is given by N. Fill value of $\frac{N}{24}$

Answer 1

3

Answer 2

The problem describes a spring-mass system subjected to an external force for a limited time, followed by free oscillation. We need to find the amplitude of the resulting simple harmonic motion (SHM).

1. Initial Equilibrium: The mass $m=0.8$ kg is suspended from a spring with spring constant $k=2000$ N/m. In equilibrium under gravity, the spring extends by $\Delta L_0$ such that $k \Delta L_0 = mg$. Let's set this equilibrium position as the origin ($y=0$) for our coordinate system, with the positive direction pointing downwards.

2. Motion under External Force ($0 \le t \le \frac{\pi}{2}$): At $t=0$, an additional downward force $F_{ext} = 72$ N starts acting. The net force on the mass at position $y$ (relative to the initial equilibrium) is the spring force $-ky$ and the external force $F_{ext}$. Gravity is already accounted for by choosing $y=0$ as the initial equilibrium position. The equation of motion is $m \frac{d^2y}{dt^2} = -ky + F_{ext}$. This can be written as $\frac{d^2y}{dt^2} + \frac{k}{m}y = \frac{F_{ext}}{m}$. The angular frequency of the natural oscillation is $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2000}{0.8}} = \sqrt{2500} = 50$ rad/s. The equation is $\frac{d^2y}{dt^2} + \omega^2 y = \frac{F_{ext}}{m}$. The general solution is $y(t) = y_h(t) + y_p(t)$, where $y_h$ is the solution to the homogeneous equation and $y_p$ is a particular solution. $y_h(t) = A \cos(\omega t + \phi)$. A particular solution is a constant shift $y_p = \frac{F_{ext}}{m \omega^2} = \frac{F_{ext}}{k}$. So, $y(t) = A \cos(\omega t + \phi) + \frac{F_{ext}}{k}$. The velocity is $v(t) = \frac{dy}{dt} = -A \omega \sin(\omega t + \phi)$.

   At $t=0$, the mass is in the initial equilibrium position ($y=0$) and at rest ($v=0$). Using $y(0)=0$: $0 = A \cos(\phi) + \frac{F_{ext}}{k} \implies A \cos(\phi) = -\frac{F_{ext}}{k}$. Using $v(0)=0$: $0 = -A \omega \sin(\phi)$. Since $A$ is the amplitude of the homogeneous solution and we expect motion ($A \ne 0$), and $\omega \ne 0$, we must have $\sin(\phi)=0$. This means $\phi = n\pi$ for some integer $n$. If $\phi = n\pi$, then $\cos(\phi) = (-1)^n$. $A (-1)^n = -\frac{F_{ext}}{k}$. We can choose $\phi=\pi$ ($n=1$), then $\cos(\phi)=-1$, giving $A(-1) = -\frac{F_{ext}}{k}$, so $A = \frac{F_{ext}}{k}$. The solution during $0 \le t \le \frac{\pi}{2}$ is $y(t) = \frac{F_{ext}}{k} \cos(\omega t + \pi) + \frac{F_{ext}}{k} = \frac{F_{ext}}{k} (1 - \cos(\omega t))$. The velocity is $v(t) = \frac{F_{ext}}{k} \omega \sin(\omega t)$.

3. State at $t = \frac{\pi}{2}$: The external force ceases to act at $t_1 = \frac{\pi}{2}$ s. We need the position and velocity at this moment. $\omega t_1 = 50 \times \frac{\pi}{2} = 25\pi$. Position at $t_1$: $y(t_1) = \frac{F_{ext}}{k} (1 - \cos(25\pi)) = \frac{F_{ext}}{k} (1 - (-1)) = 2 \frac{F_{ext}}{k}$. $y(t_1) = 2 \times \frac{72}{2000} = 2 \times 0.036 = 0.072$ m. Velocity at $t_1$: $v(t_1) = \frac{F_{ext}}{k} \omega \sin(25\pi) = \frac{F_{ext}}{k} \omega (0) = 0$.

4. Resulting SHM (for $t > \frac{\pi}{2}$): After $t = \frac{\pi}{2}$, the external force is removed. The equation of motion becomes $m \frac{d^2y}{dt^2} = -ky$. This is SHM about the initial equilibrium position ($y=0$). The motion is described by $y(t) = A' \cos(\omega (t - t_1) + \phi')$, where $A'$ is the amplitude of the resulting SHM and $\omega=50$ rad/s. Let $\tau = t - t_1$. The initial conditions for this SHM at $\tau=0$ are the state at $t_1$: $y(\tau=0) = y(t_1) = 0.072$ m. $v(\tau=0) = v(t_1) = 0$.

   Using $y(0) = 0.072$: $0.072 = A' \cos(\phi')$. Using $v(0) = 0$: $v(\tau) = -A' \omega \sin(\omega \tau + \phi')$, so $0 = -A' \omega \sin(\phi')$. Since $A' \ne 0$ (as $y(0) \ne 0$) and $\omega \ne 0$, we must have $\sin(\phi')=0$. This means $\phi' = m\pi$ for some integer $m$. $0.072 = A' \cos(m\pi) = A' (-1)^m$. Since amplitude $A'$ is positive, we choose $m=0$, so $\phi'=0$. Then $0.072 = A' (1)$, which gives $A' = 0.072$ m.

5. Amplitude and Final Calculation: The amplitude of the resulting SHM is $A' = 0.072$ m. The question asks for the amplitude in mm, which is $N$. $N = 0.072 \times 1000$ mm = 72 mm. We need to find the value of $\frac{N}{24}$. $\frac{N}{24} = \frac{72}{24} = 3$.

In this case, at $t=\pi/2$, the position is $y_0 = 0.072$ m and the velocity is $v_0 = 0$. The angular frequency is $\omega=50$ rad/s. The amplitude is $A' = \sqrt{(0.072)^2 + (0/50)^2} = \sqrt{(0.072)^2} = 0.072$ m. $N = 0.072$ m = 72 mm. $\frac{N}{24} = \frac{72}{24} = 3$.