/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 The period of oscillation of a m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 3

The period of oscillation of a mass \(m\) suspended from a spring is \(2 \mathrm{~s}\). If along with it another mass \(2 \mathrm{~kg}\) is also suspended, the period of oscillation increases by \(1 \mathrm{~s} .\) The mass \(m\) will be (a) \(2 \mathrm{~kg}\) (b) \(1 \mathrm{~kg}\) (c) \(1.6 \mathrm{~kg}\) (d) \(2.6 \mathrm{~kg}\)

Short Answer

Expert verified
The mass \( m \) is \( 1.6 \text{ kg} \).

Step by step solution

01

Understanding the Formula for Period of Oscillation

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

Calculate Spring Constant Using Initial Conditions

Given that the period \( T = 2 \text{ s} \) initially, we can write \( 2 = 2\pi \sqrt{\frac{m}{k}} \). Solving for \( k \), we get \( k = \frac{\pi^2 m}{1} = \pi^2 m \).
03

Calculate New Period with Additional Mass

With an additional mass of \( 2 \text{ kg} \), the total mass becomes \( m + 2 \). The new period is \( 3 \text{ s} \), thus \( 3 = 2\pi \sqrt{\frac{m + 2}{k}} \).
04

Equate and Solve for Mass

Using the two equations for \( k \), equate \( k = \pi^2 m \) from the first condition and \( k = 4\pi^2 \times \frac{m+2}{9} \) from the second condition. Set them equal: \( \pi^2 m = 4\pi^2 \times \frac{m+2}{9} \).
05

Simplify and Solve the Equation

Simplify to \( m = \frac{4(m+2)}{9} \). Multiply through by 9 to get \( 9m = 4(m+2) \). Expand and simplify: \( 9m = 4m + 8 \). Subtract \( 4m \) from both sides to find \( 5m = 8 \). Divide by 5: \( m = \frac{8}{5} \).
06

Finalize and Find the Correct Option

Thus \( m = 1.6 \text{ kg} \). Therefore, the correct answer choice is (c) \( 1.6 \text{ kg} \).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass-Spring System
The mass-spring system is a fundamental concept in physics that demonstrates simple harmonic motion. In this system, a block or mass is attached to the end of a spring. When the mass is displaced from its equilibrium position (either stretched or compressed), it experiences a force from the spring, which is directly proportional to the displacement. This force, according to Hooke's Law, is given by:\[ F = -kx \]* Here, \( F \) is the restoring force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement of the mass from its equilibrium. In a mass-spring system, the mass moves back and forth about an equilibrium point, which is an example of simple harmonic motion. Understanding how the mass-spring system operates is crucial for analyzing any oscillatory motion involved in such a setup.
Spring Constant
The spring constant, denoted by \( k \), is a measure of the stiffness of a spring. A higher \( k \) value indicates a stiffer spring, while a smaller value represents a more flexible one. It plays a crucial role in determining the behavior of a mass-spring system.* The unit for the spring constant is Newton per meter (N/m).The relationship between period \( T \), mass \( m \), and spring constant \( k \) in a mass-spring system is expressed by the equation:\[ T = 2\pi \sqrt{\frac{m}{k}} \]In the context of the exercise, the spring constant affects the period of oscillation. By knowing this constant, one can predict how faster or slower the system will oscillate based on the mass attached to the spring. The ability to calculate \( k \) from given conditions (as shown in the exercise) is crucial for solving oscillation problems.
Simple Harmonic Motion
Simple harmonic motion (SHM) is a type of periodic motion where the restoring force is proportional to the displacement. In the mass-spring system, SHM occurs when the mass oscillates due to the restoring force from the spring.Some key features of simple harmonic motion include:
  • The motion is sinusoidal in nature.
  • The system exhibits a constant period, unaffected by amplitude.
  • Energy is conserved, alternating between kinetic and potential forms.
The equation for SHM in a mass-spring system is given by the formula:\[ x(t) = A \cos(\omega t + \phi) \]* \( x(t) \) is the displacement at time \( t \), \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase angle.Understanding SHM is essential for predicting the movement of the mass-spring system, as it describes how the system's position changes over time during oscillation. In practice, calculations relating to SHM often involve determining the period, frequency, and amplitude of oscillations in physical systems.

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

A weightless spring which has a force constant \(k\) oscillates with frequency \(n\) when a mass \(m\) is suspended from it. The spring is cut into two equal halves and a mass \(2 \mathrm{~m}\) is suspended from one part of spring. The frequency of oscillation will now become (a) \(n\) (b) \(2 n\) (c) \(\frac{n}{\sqrt{2}}\) (d) \(n(2)^{1 / 2}\)

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force, \(F=F_{0} \sin \omega t\) If, the amplitude of the particle is maximum for \(\omega=\omega_{1}\) and the energy of the particle is maximum for \(\omega=\omega_{2}\), then (a) \(\omega_{1}=\omega_{0}\) and \(\omega_{2} \neq \omega_{0}\) (b) \(\omega_{1}=\omega_{0}\) and \(\omega_{2}=\omega_{0}\) (c) \(\omega_{1} \neq \omega_{0}\) and \(\omega_{2}=\omega_{0}\) (d) \(\omega_{1} \neq \omega_{0}\) and \(\omega_{2} \neq \omega_{0}\)

When a body of mass \(1.0 \mathrm{~kg}\) is suspended from a certain light spring hanging vertically, its length increases by \(5 \mathrm{~cm}\). By suspending \(2.0 \mathrm{~kg}\) block to the spring and if the block is pulled through \(10 \mathrm{~cm}\) and released, the maximum velocity of it, in \(\mathrm{ms}^{-1}\) is \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(0.5\) (b) 1 (c) 2 (d) 4

On a smooth inclined plane, a body of mass \(M\) is attached between two springs. The other ends of the springs are fixed to firm support. If each spring has force constant \(k\), the period of oscillation of the body (assuming the springs as massless) is (a) \(2 \pi[M / 2 k]^{1 / 2}\) (b) \(2 \pi[2 M / k]^{1 / 2}\) (c) \(2 \pi[M g \sin \theta / 2 k]^{1 / 2}\) (d) \(2 \pi[2 M g / k]^{1 / 2}\)

Two pendulums of length \(1 \mathrm{~m}\) and \(16 \mathrm{~m}\) start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is \(T\). The minimum time after which the two threads of the pendulum will be one behind the other is (a) \(T / 4\) (b) \(T / 3\) \(\begin{array}{ll}\text { (c) } 4 T / 3 & \text { (d) } 4 T\end{array}\)
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Work Energy and Power

Read Explanation

Radiation

Read Explanation

Particle Model of Matter

Read Explanation

Torque and Rotational Motion

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.