Question:
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 ______
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 ______
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Hint:
Write Newton's second law for both blocks, assume \(x_i \propto e^{i\omega t}\), and set the determinant of the resulting \(2\times2\) system to zero to get \(\omega^4 - 16\omega^2 + 32 = 0\).
Write Newton's second law for both blocks, assume \(x_i \propto e^{i\omega t}\), and set the determinant of the resulting \(2\times2\) system to zero to get \(\omega^4 - 16\omega^2 + 32 = 0\).
Updated On: Jul 28, 2026
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Correct Answer: 2.41
Solution and Explanation
Step 1: Understanding the Concept:
Set up an x-axis along the line of motion. Let \(x_1\) and \(x_2\) be the displacements of block 1 and block 2 from their equilibrium positions. The first spring (\(k_1 = 8\) N/m) connects the wall to block 1, and the second spring (\(k_2 = 4\) N/m) connects block 1 to block 2. Each block has mass \(m = 1\) kg.
Step 2: Write Newton's second law for each block.
Block 1 feels a restoring force from spring 1 (stretched by \(x_1\)) and from spring 2 (stretched by \(x_2 - x_1\)):
\[ m\ddot{x}_1 = -k_1 x_1 + k_2(x_2 - x_1) = -(k_1+k_2)x_1 + k_2 x_2 \]
Block 2 only feels the second spring:
\[ m\ddot{x}_2 = -k_2(x_2 - x_1) = k_2 x_1 - k_2 x_2 \]
With \(m=1\), \(k_1=8\), \(k_2=4\):
\[ \ddot{x}_1 = -12x_1 + 4x_2, \qquad \ddot{x}_2 = 4x_1 - 4x_2 \]
Step 3: Find the normal mode frequencies.
Assume \(x_1, x_2 \propto e^{i\omega t}\), so \(\ddot x_i \to -\omega^2 x_i\). Substituting turns the equations into:
\[ (12-\omega^2)x_1 = 4x_2, \qquad 4x_1 = (4-\omega^2)x_2 \]
For a non-trivial solution the determinant of the coefficients must vanish:
\[ (12-\omega^2)(4-\omega^2) - 16 = 0 \]
\[ \omega^4 - 16\omega^2 + 32 = 0 \]
Step 4: Solve the quadratic in \(\omega^2\).
\[ \omega^2 = \frac{16 \pm \sqrt{256-128}}{2} = 8 \pm 4\sqrt{2} \]
So \(\omega_H^2 = 8+4\sqrt{2} = 13.66\) and \(\omega_L^2 = 8-4\sqrt{2} = 2.34\), giving \(\omega_H = 3.70\) rad/s and \(\omega_L = 1.53\) rad/s.
Final Answer:
The ratio simplifies neatly: \(\dfrac{\omega_H}{\omega_L} = \sqrt{\dfrac{8+4\sqrt2}{8-4\sqrt2}} = 1+\sqrt2 = 2.41\). \[ \boxed{\dfrac{\omega_H}{\omega_L} = 2.41} \]
Set up an x-axis along the line of motion. Let \(x_1\) and \(x_2\) be the displacements of block 1 and block 2 from their equilibrium positions. The first spring (\(k_1 = 8\) N/m) connects the wall to block 1, and the second spring (\(k_2 = 4\) N/m) connects block 1 to block 2. Each block has mass \(m = 1\) kg.
Step 2: Write Newton's second law for each block.
Block 1 feels a restoring force from spring 1 (stretched by \(x_1\)) and from spring 2 (stretched by \(x_2 - x_1\)):
\[ m\ddot{x}_1 = -k_1 x_1 + k_2(x_2 - x_1) = -(k_1+k_2)x_1 + k_2 x_2 \]
Block 2 only feels the second spring:
\[ m\ddot{x}_2 = -k_2(x_2 - x_1) = k_2 x_1 - k_2 x_2 \]
With \(m=1\), \(k_1=8\), \(k_2=4\):
\[ \ddot{x}_1 = -12x_1 + 4x_2, \qquad \ddot{x}_2 = 4x_1 - 4x_2 \]
Step 3: Find the normal mode frequencies.
Assume \(x_1, x_2 \propto e^{i\omega t}\), so \(\ddot x_i \to -\omega^2 x_i\). Substituting turns the equations into:
\[ (12-\omega^2)x_1 = 4x_2, \qquad 4x_1 = (4-\omega^2)x_2 \]
For a non-trivial solution the determinant of the coefficients must vanish:
\[ (12-\omega^2)(4-\omega^2) - 16 = 0 \]
\[ \omega^4 - 16\omega^2 + 32 = 0 \]
Step 4: Solve the quadratic in \(\omega^2\).
\[ \omega^2 = \frac{16 \pm \sqrt{256-128}}{2} = 8 \pm 4\sqrt{2} \]
So \(\omega_H^2 = 8+4\sqrt{2} = 13.66\) and \(\omega_L^2 = 8-4\sqrt{2} = 2.34\), giving \(\omega_H = 3.70\) rad/s and \(\omega_L = 1.53\) rad/s.
Final Answer:
The ratio simplifies neatly: \(\dfrac{\omega_H}{\omega_L} = \sqrt{\dfrac{8+4\sqrt2}{8-4\sqrt2}} = 1+\sqrt2 = 2.41\). \[ \boxed{\dfrac{\omega_H}{\omega_L} = 2.41} \]
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