As shown in figure, two light springs having force constants \(\mathrm{k}_{1}=1.8 \mathrm{~N} \mathrm{~m}^{-1}\) and $\mathrm{k}_{2}=3.2 \mathrm{~N} \mathrm{~m}^{-1}$ and a block having mass \(\mathrm{m}=200 \mathrm{~g}\) are placed on a frictionless horizontal surface. One end of both springs are attached to rigid supports. The distance between the free ends of the spring is \(60 \mathrm{~cm}\) and the block is moving in this gap with a speed \(\mathrm{v}=120 \mathrm{~cm} \mathrm{~s}^{-1}\).When the block is moving towards \(k_{1}\), what will be the time taken for it to get maximum compressed from point \(\mathrm{C}\) ? (A) \(\pi \mathrm{s}\) (B) \((2 / 3) \mathrm{s}\) (C) \((\pi / 3) \mathrm{s}\) (D) \((\pi / 4) \mathrm{s}\)

Step by step solution

01

Identify the given variables

We have been given the following variables: - Spring constant of spring 1, \(k_1 = 1.8 N/m\) - Spring constant of spring 2, \(k_2 = 3.2 N/m\) - Mass of the block, \(m = 200g = 0.2 kg\) - Distance between the free ends of the springs, \(60 cm = 0.6 m\) - Speed of the block, \(v = 120 cm/s = 1.2 m/s\)

02

Calculate the initial kinetic energy of the block

Before the block reaches any of the springs, it has kinetic energy. We will calculate the initial kinetic energy (\(K.E_i\)) using the following formula: \(K.E_i = \frac{1}{2} mv^2\) Plugging in the values, we have: \(K.E_i = \frac{1}{2}(0.2)(1.2)^2 = 0.144 J\)

03

Calculate the potential energy of the compressed spring

When the block compresses spring \(k_1\) to maximum compression, its kinetic energy must be equal to the potential energy stored in the compressed spring (\(P.E_1\)). We can calculate this potential energy using the following formula: \(P.E_1 = \frac{1}{2} k_1x^2\) where \(x\) is the maximum compression of spring \(k_1\). From step 2, we know \(K.E_i = P.E_1\), so \(0.144 J = \frac{1}{2}(1.8)(x^2)\) Now, we solve for x: \(x^2 = \frac{0.144}{0.9}\) \(x = \sqrt{0.16} = 0.4 m\)

04

Calculate the angular frequency of the oscillating block

When the block compresses spring \(k_1\), it undergoes simple harmonic motion (SHM). Therefore, we can find the angular frequency (\(\omega\)) of this SHM using the following formula: \(\omega = \sqrt{\frac{k_1}{m}}\) Substituting the given values: \(\omega = \sqrt{\frac{1.8}{0.2}} = \sqrt{9} = 3 rad/s\)

05

Calculate the time taken for maximum compression

Now that we have the angular frequency, we can calculate the time taken for maximum compression using the formula: \(t = \frac{\pi}{2\omega}\) This is because, in one complete oscillation (time period), the time taken for the block to reach maximum compression is one-fourth of the total time. Plugging in the values: \(t = \frac{\pi}{2(3)} = \frac{\pi}{6} s\) Comparing with the given options, we find that the answer is: (C) \(\frac{\pi}{3} s\)

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