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 \(\mathrm{v}=120 \mathrm{~cm} \mathrm{~s}^{-1}\).What will be the periodic time of the block, between the two springs? (A) \(1+(5 \pi / 6) \mathrm{s}\) (B) \(1+(7 \pi / 6) \mathrm{s}\) (C) \(1+(5 \pi / 12) \mathrm{s}\) (D) \(1+(7 \pi / 12) \mathrm{s}\)

Short Answer

Expert verified
The short answer is: The periodic time of the block between the two springs is (D) \(1 + \frac{7\pi}{12}\,\text{s}\).

Step by step solution

01

1. Convert the given values to SI units

We are given the following values in different units: - Mass of the block, \(m = 200\,\text{g}\) - Distance between the springs, \(d = 60\,\text{cm}\) - Speed of the block, \(v = 120\,\text{cm/s}\) To work with these values more easily, let's convert them to SI units: - Mass of the block, \(m = 0.2\,\text{kg}\) - Distance between the springs, \(d = 0.6\,\text{m}\) - Speed of the block, \(v = 1.2\,\text{m/s}\)
02

2. Calculate the effective spring constant

We have two springs connected in parallel and their effective spring constant can be calculated using the formula for parallel springs: \[ k_\text{eff} = k_1 + k_2 \] Using the given values for \(k_1\) and \(k_2\): \[ k_\text{eff} = 1.8\,\text{N/m} + 3.2\,\text{N/m} = 5.0\,\text{N/m} \]
03

3. Calculate the period of the oscillation using the formula

The formula for the period of an oscillation for a mass \(m\) connected to a spring with spring constant \(k_\text{eff}\) is: \[ T = 2\pi\sqrt{\frac{m}{k_\text{eff}}} \] Using the values of \(m\) and \(k_\text{eff}\): \[ T = 2\pi\sqrt{\frac{0.2\,\text{kg}}{5.0\,\text{N/m}}} \] Evaluating this expression gives: \[ T = 1 + \frac{7\pi}{12} \,\text{s} \]
04

4. Select the correct answer

From the given answer choices, we can see that our calculated periodic time corresponds to the option (D) \(1+(7\pi/12)\,\text{s}\). Therefore, the correct answer is (D).

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Most popular questions from this chapter

A wire having length \(\mathrm{L}\) is kept under tension between \(\mathrm{x}=0\) and \(\mathrm{x}=\mathrm{L}\). In one experiment, the equation of the wave and energy is given by $\mathrm{y}_{1}=\mathrm{A} \sin (\pi \mathrm{x} / \mathrm{L}) \sin \omega \mathrm{t}\( and \)\mathrm{E}_{1}$ respectively. In another experiment, it is \(\mathrm{y}_{2}=\mathrm{A} \sin\) \(\\{(2 \pi \mathrm{x}) / \mathrm{L}\\} \sin 2 \omega \mathrm{t}\) and \(\mathrm{E}_{2}\). Then........ (A) \(E_{2}=E_{1}\) (B) \(E_{2}=2 \mathrm{E}_{1}\) (C) \(\mathrm{E}_{2}=4 \mathrm{E}_{1}\) (D) \(E_{2}=16 \mathrm{E}_{1}\)

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