Two 1 kg blocks are connected to two massless springs of spring constants $8\text{ N.m}^{-1}$ and $4\text{ N.m}^{-1}$. The system is kept on a frictionless horizontal floor with one end of a spring attached to a wall (see figure below). They are performing oscillatory motion along the x-axis with the normal mode frequencies $\omega_H$ and $\omega_L$ ($\omega_H > \omega_L$). The ratio $\dfrac{\omega_H}{\omega_L}$ (rounded off to two decimal places) is ______

Question

On a horizontal plane, a projectile of mass $m$ is launched from the ground with speed $v_0$ at an angle $\theta_0$ with the horizontal. In addition to the gravitational force ($mg$), it also experiences a drag force $\vec{F}_{drag} = -\gamma \vec{v}$, where $\vec{v}$ is its velocity and $\gamma$ is a constant. It hits the ground at a distance $R$ from the point of launch with its velocity making an angle $\theta$ with the horizontal, as shown schematically in the figure below.

Then which of the following options is correct?

Answer 1

Step 1: Write the kinetic and potential energy.
Take $x_1$, $x_2$ as the displacements of the two blocks from equilibrium. The kinetic energy of the system is:
\[ T = \frac{1}{2}m\dot x_1^2 + \frac{1}{2}m\dot x_2^2 \]
The potential energy stored in the two springs (wall-to-block-1, and block-1-to-block-2) is:
\[ U = \frac{1}{2}k_1 x_1^2 + \frac{1}{2}k_2(x_2-x_1)^2 \]

Step 2: Get the equations of motion from the Lagrangian.
With $L = T-U$, the Euler-Lagrange equations $\frac{d}{dt}\frac{\partial L}{\partial \dot x_i} - \frac{\partial L}{\partial x_i}=0$ give:
\[ m\ddot x_1 + (k_1+k_2)x_1 - k_2 x_2 = 0, \qquad m\ddot x_2 - k_2 x_1 + k_2 x_2 = 0 \]
This is the same as writing $M\ddot{\mathbf{x}} = -K\mathbf{x}$ with mass matrix $M = mI$ and stiffness matrix $K = \begin{pmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2\end{pmatrix}$.

Step 3: Solve the generalized eigenvalue problem.
Normal modes need $\det(K - \omega^2 M) = 0$. With $m=1$, $k_1=8$, $k_2=4$:
\[ \det\begin{pmatrix} 12-\omega^2 & -4 \\ -4 & 4-\omega^2 \end{pmatrix} = 0 \]
\[ (12-\omega^2)(4-\omega^2) - 16 = 0 \implies \omega^4 - 16\omega^2 + 32 = 0 \]

Step 4: Solve for the two roots.
\[ \omega^2 = 8 \pm 4\sqrt{2} \]
so $\omega_H = \sqrt{8+4\sqrt2} \approx 3.70$ rad/s and $\omega_L = \sqrt{8-4\sqrt2} \approx 1.53$ rad/s.

Final Answer:
The ratio is $\omega_H/\omega_L = \sqrt{(8+4\sqrt2)/(8-4\sqrt2)} = 1+\sqrt2 \approx 2.41$. \[ \boxed{\dfrac{\omega_H}{\omega_L} = 2.41} \]

Show Hint

Correct Answer: 2.41

Solution and Explanation

Top Questions on Classical Mechanics

Questions Asked in GATE PH exam