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Chapter 10: Problem 1383

When a mass \(\mathrm{m}\) is suspended from the free end of a massless spring having force constant \(\mathrm{k}\), its oscillates with frequency \(\mathrm{f}\). Now if the spring is divided into two equal parts and a mass $2 \mathrm{~m}$ is suspended from the end of anyone of them, it will oscillate with a frequency equal to......... (A) \(\mathrm{f}\) (B) \(2 \mathrm{f}\) (C) \((\mathrm{f} / \sqrt{2})\) (D) \(\sqrt{2 f}\)

Short Answer

Expert verified
The frequency of oscillation when a mass $2\mathrm{m}$ is suspended from the end of one of the divided springs is equal to the initial frequency, $f$. So, the answer is (A) \(\mathrm{f}\).

Step by step solution

01

Expression for frequency of oscillation for mass m

We can first express the frequency of oscillation for a mass m suspended from a spring with force constant k using the formula: \(f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\)
02

Force constant of the divided spring

When the spring is divided into two equal parts, its force constant changes. The force constant for each part of the spring when it is divided is given by: \(k' = 2k\) The new force constant, \(k'\), is twice the original force constant.
03

Expression for frequency of oscillation for mass 2m and divided spring

Now we need to find the frequency of oscillation when a mass 2m is suspended from the end of one of the divided springs having force constant \(k'\). We can use the same formula for frequency, with the new values for mass and force constant: \(f' = \dfrac{1}{2\pi}\sqrt{\dfrac{k'}{2m}}\)
04

Substitute the new force constant

We can now substitute the new force constant, \(k' = 2k\), into the expression for frequency: \(f' = \dfrac{1}{2\pi}\sqrt{\dfrac{2k}{(2m)}}\)
05

Simplify and compare with the original frequency

Simplify the expression for the new frequency: \(f' = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\) Comparing this expression with the original frequency expression, we can see that they are equal: \(f' = f\) Therefore, the correct answer is: (A) \(\mathrm{f}\)

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Most popular questions from this chapter

As shown in figure, a spring attached to the ground vertically has a horizontal massless plate with a \(2 \mathrm{~kg}\) mass in it. When the spring (massless) is pressed slightly and released, the \(2 \mathrm{~kg}\) mass, starts executing S.H.M. The force constant of the spring is \(200 \mathrm{Nm}^{-1}\). For what minimum value of amplitude, will the mass loose contact with the plate? (Take \(\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(10.0 \mathrm{~cm}\) (B) \(8.0 \mathrm{~cm}\) (C) \(4.0 \mathrm{~cm}\) (D) For any value less than \(12.0 \mathrm{~cm}\).

The displacement of a S.H.O. is given by the equation \(\mathrm{x}=\mathrm{A}\) \(\cos \\{\omega t+(\pi / 8)\\}\). At what time will it attain maximum velocity? (A) \((3 \pi / 8 \omega)\) (B) \((8 \pi / 3 \omega)\) (C) \((3 \pi / 16 \omega)\) (D) \((\pi / 16 \pi)\).

If the maximum frequency of a sound wave at room temperature is $20,000 \mathrm{~Hz}\( then its minimum wavelength will be approximately \)\ldots \ldots\left(\mathrm{v}=340 \mathrm{~ms}^{-1}\right)$ (A) \(0.2 \AA\) (B) \(5 \AA\) (C) \(5 \mathrm{~cm}\) to \(2 \mathrm{~m}\) (D) \(20 \mathrm{~mm}\)

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