Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 1418

A block having mass \(\mathrm{M}\) is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant \(\mathrm{k}\). The other end of the spring is attached to a rigid wall. This system consisting of spring and mass \(\mathrm{M}\) is executing SHM with amplitude \(\mathrm{A}\) and frequency \(\mathrm{f}\). When the block is passing through the mid-point of its path of motion, a body of mass \(\mathrm{m}\) is placed on top of it, as a result of which its amplitude and frequency changes to \(\mathrm{A}^{\prime}\) and \(\mathrm{f}\). The ratio of frequencies \((\mathrm{f} / \mathrm{f})=\ldots \ldots \ldots\) (A) \(\sqrt{\\{} \mathrm{M} /(\mathrm{m}+\mathrm{M})\\}\) (B) \(\sqrt{\\{\mathrm{m} /(\mathrm{m}+\mathrm{M})\\}}\) (C) \(\sqrt{\\{\mathrm{MA} / \mathrm{mA}}\\}\) (D) \(\sqrt{[}\\{(\mathrm{M}+\mathrm{m}) \mathrm{A}\\} / \mathrm{mA}]\)

Short Answer

Expert verified
None of the given options are correct. The ratio of the initial frequency to the final frequency is: \(\frac{f}{f'} = \sqrt{1 + \frac{m}{M}}\)

Step by step solution

01

Frequency of Simple Harmonic Motion

The frequency of a mass-spring system undergoing simple harmonic motion is given by the following formula: \(f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}\) Initially, the frequency of the block with mass M is: \(f = \frac{1}{2 \pi} \sqrt{\frac{k}{M}}\)
02

Frequency of the Combined System

When mass m is added to the system, the total mass becomes (M + m), and the frequency of the combined system is given by: \(f' = \frac{1}{2 \pi} \sqrt{\frac{k}{M + m}}\)
03

Calculate the Ratio of Frequencies

Now, we want to find the ratio of the initial frequency (f) to the final frequency (f'): \(\frac{f}{f'} = \frac{\frac{1}{2 \pi} \sqrt{\frac{k}{M}}}{\frac{1}{2 \pi} \sqrt{\frac{k}{M + m}}}\)
04

Simplify the Ratio

We can now simplify this expression, canceling out common terms: \(\frac{f}{f'} = \frac{\sqrt{\frac{k}{M}}}{\sqrt{\frac{k}{M + m}}}\) Dividing two square roots is the same as taking the square root of the ratio: \(\frac{f}{f'} = \sqrt{\frac{\frac{k}{M}}{\frac{k}{M + m}}}\) Cancel out k in the numerator and denominator: \(\frac{f}{f'} = \sqrt{\frac{M + m}{M}}\) Which is equivalent to: \(\frac{f}{f'} = \sqrt{1 + \frac{m}{M}}\)
05

Choosing the Correct Option

Comparing this to the given options, we find that none of the options exactly match our final answer. It's important to double check our working and ensure we didn't miss any steps, but assuming we followed the correct process, it looks like there could be an error in the question or the given options.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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) \(1+(7 \pi / 12) \mathrm{s}\)

Two waves are represented by $\mathrm{y}_{1}=\mathrm{A} \sin \omega \mathrm{t}\( and \)\mathrm{y}_{2}=\mathrm{A}\( cos \)\omega \mathrm{t}$. The phase of the first wave, \(\mathrm{w}\). r. t. to the second wave is (A) more by radian (B) less by \(\pi\) radian (C) more by \(\pi / 2\) (D) less by \(\pi / 2\)

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The frequency of the particle is \(\ldots \ldots \mathrm{s}^{-1}\). (A) \((1 / \pi)\) (B) \(\pi\) (C) \((1 / 2 \pi)\) (D) \((\pi / 2)\)

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) \((\pi / 16 \pi)\).
See all solutions

Recommended explanations on English Textbooks

Semiotics

Read Explanation

Prosody

Read Explanation

Essay Prompts

Read Explanation

Multiple Choice Questions

Read Explanation

Graphology

Read Explanation

Sociolinguistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free