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Chapter 10: Problem 1366

When a mass \(M\) is suspended from the free end of a spring, its periodic time is found to be \(\mathrm{T}\). Now, if the spring is divided into two equal parts and the same mass \(\mathrm{M}\) is suspended and oscillated, the periodic time of oscillation is found to be \(\mathrm{T}\) '. Then \(\ldots \ldots \ldots\) (A) \(\mathrm{T}<\mathrm{T}^{\prime}\) (B) \(\mathrm{T}=\mathrm{T}^{\prime}\) (C) \(\mathrm{T}>\mathrm{T}^{\prime}\) (D) Nothing can be said.

Short Answer

Expert verified
(A) \(T < T'\)

Step by step solution

01

Write down the formula for the periodic time

The formula for the periodic time of a mass-spring system is given by \(T = 2\pi\sqrt{\frac{M}{k}}\) where \(T\) is the periodic time, \(M\) is the mass, and \(k\) is the spring constant of the spring.
02

Determine the relation between periodic time and spring constant

We are given that when the spring is divided into two equal parts, the mass M will oscillate with a new periodic time T'. Let's denote the spring constant of the original undivided spring as \(k_1\) and the spring constant of the divided springs as \(k_2\). We will use the above formula for the periodic time with these two spring constants to obtain: \(T = 2\pi\sqrt{\frac{M}{k_1}}\) and \(T' = 2\pi\sqrt{\frac{M}{k_2}}\)
03

Relate the spring constants of the divided and undivided spring

We know that when we divide the spring into two equal parts, the spring constant is also changed. For a spring, the spring constant is inversely proportional to its length (\(k \propto \frac{1}{L}\)). Since the spring is divided into two equal parts, the length of each part is half the length of the original spring. Thus, the spring constant of the divided springs, \(k_2\), is twice the spring constant of the original spring: \(k_2 = 2k_1\)
04

Compare the periodic times T and T'

Now we can use this relationship between the spring constants to compare the periodic times T and T'. By plugging the expression for \(k_2\) in terms of \(k_1\) from Step 3 into the T' equation from Step 2, we can obtain: \(T' = 2\pi\sqrt{\frac{M}{2k_1}}\) Now, dividing the T' equation by the original T equation: \(\frac{T'}{T} = \frac{2\pi\sqrt{\frac{M}{2k_1}}}{2\pi\sqrt{\frac{M}{k_1}}} = \sqrt{\frac{1}{2}}\) Since \(0 < \sqrt{\frac{1}{2}} < 1\), we have: \(T' < T\) So, the correct answer is: (A) \(T < T'\)

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Most popular questions from this chapter

The periodic time of a S.H.O. oscillating about a fixed point is $2 \mathrm{~s}$. After what time will the kinetic energy of the oscillator become \(25 \%\) of its total energy? (A) \(1 / 12 \mathrm{~s}\) (B) \(1 / 6 \mathrm{~s}\) (C) \(1 / 4 \mathrm{~s}\) (D) \(1 / 3 \mathrm{~s}\).

A rectangular block having mass \(\mathrm{m}\) and cross sectional area A is floating in a liquid having density \(\rho\). If this block in its equilibrium position is given a small vertical displacement, its starts oscillating with periodic time \(\mathrm{T}\). Then in this case \(\ldots \ldots\) (A) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{m}})\) (B) \(T \propto \sqrt{\rho}\) (C) \(\mathrm{T} \propto(1 / \sqrt{\mathrm{A}})\) (D) \(\mathrm{T} \propto(1 / \sqrt{\rho})\)

Tube \(A\) has both ends open while tube \(B\) has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube \(\mathrm{A}\) and \(\mathrm{B}\) is \(\ldots \ldots \ldots\) (A) \(1: 2\) (B) \(1: 4\) (C) \(2: 1\) (D) \(4: 1\)

The ratio of force constants of two springs is \(1: 5\). The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is \(\ldots \ldots\) (A) \((1 / \sqrt{5})\) (B) \((1 / 5)\) (C) \((5 / 1)\) (D) \((\sqrt{5} / 1)\)

If two SHM's are given by the equation $\mathrm{y}_{1}=0.1 \sin [\pi \mathrm{t}+(\pi / 3)]\( and \)\mathrm{y}_{2}=0.1 \cos \pi \mathrm{t}$, then the phase difference between the velocity of particle 1 and 2 is \(\ldots \ldots \ldots\) (A) \(\pi / 6\) (B) \(-\pi / 3\) (C) \(\pi / 3\) (D) \(-\pi / 6\)
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