An empty cart, tied between two ideal springs, oscillates with $\omega=10.0 \mathrm{rad} / \mathrm{s} .$ A load is placed in the cart. making the total mass 4.0 times what it was before. What is the new value of \(\omega ?\)

Short Answer

Expert verified
When the total mass of the oscillating cart increases to 4 times its original value, the new angular frequency is 5.0 rad/s.

Step by step solution

01

Understand the given information

We are given: 1. The initial angular frequency, \(\omega = 10.0\, \mathrm{rad} / \mathrm{s}\). 2. The mass becomes 4 times its original value after placing a load. We need to find the new value of \(\omega\) after the mass change.
02

Write the formula for the angular frequency

The formula for the angular frequency of a mass-spring system is: \(\omega = \sqrt{\frac{k}{m}}\) In our case, \(k\) (the spring constant) does not change, but the mass (\(m\)) does.
03

Find the initial mass

Before we can find the new value of \(\omega\), let's assume the initial mass is \(m_i\), so we can write the initial angular frequency as: \(\omega_i = \sqrt{\frac{k}{m_i}}\) Given the initial angular frequency as \(\omega_i = 10.0\, \mathrm{rad} / \mathrm{s}\), we can rewrite the previous equation as: \(10.0 = \sqrt{\frac{k}{m_i}}\)
04

Find the new mass

The new mass is 4 times the initial mass, so we can express it as follows: \(m_n = 4m_i\)
05

Write the formula for the new angular frequency

Now, we need to find the new angular frequency (\(\omega_n\)) when the mass is \(m_n\). We can write this new equation as: \(\omega_n = \sqrt{\frac{k}{m_n}}\)
06

Find the relationship between the initial and new angular frequency

We can find a relationship between the initial and new angular frequency by dividing the equation for \(\omega_n\) by the equation for \(\omega_i\): \(\frac{\omega_n}{\omega_i} = \sqrt{\frac{k/m_n}{k/m_i}}\) Since the spring constant \(k\) is the same for both cases, we can cancel them out: \(\frac{\omega_n}{\omega_i} = \sqrt{\frac{m_i}{m_n}}\)
07

Calculate the new angular frequency

Using the relationship found in the previous step, we can calculate the new angular frequency (\(\omega_n\)). We know that the initial angular frequency is \(10.0\, \mathrm{rad} / \mathrm{s}\), and the new mass (\(m_n = 4m_i\)). Therefore, we can rewrite our relationship as: \(\frac{\omega_n}{10.0} = \sqrt{\frac{m_i}{4m_i}}\) Now, solve for \(\omega_n\): \(\omega_n = 10.0 \cdot \sqrt{\frac{m_i}{4m_i}}\) As the initial mass \(m_i\) appears in both the numerator and the denominator, it cancels out: \(\omega_n = 10.0 \cdot \sqrt{\frac{1}{4}}\) Solving for the new angular frequency: \(\omega_n = 10.0 \cdot \frac{1}{2} = 5.0\, \mathrm{rad} / \mathrm{s}\) The new value of \(\omega\) is 5.0 rad/s when the total mass becomes 4 times its original value.

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 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

A pendulum (mass \(m\), unknown length) moves according to \(x=A\) sin $\omega t .\( (a) Write the equation for \)v_{x}(t)$ and sketch one cycle of the \(v_{x}(t)\) graph. (b) What is the maximum kinetic energy?
A marble column with a cross-sectional area of \(25 \mathrm{cm}^{2}\) supports a load of \(7.0 \times 10^{4} \mathrm{N} .\) The marble has a Young's modulus of \(6.0 \times 10^{10} \mathrm{Pa}\) and a compressive strength of $2.0 \times 10^{8} \mathrm{Pa} .$ (a) What is the stress in the column? (b) What is the strain in the column? (c) If the column is \(2.0 \mathrm{m}\) high, how much is its length changed by supporting the load? (d) What is the maximum weight the column can support?
The leg bone (femur) breaks under a compressive force of about $5 \times 10^{4} \mathrm{N}\( for a human and \)10 \times 10^{4} \mathrm{N}$ for a horse. The human femur has a compressive strength of \(1.6 \times 10^{8} \mathrm{Pa},\) while the horse femur has a compressive strength of $1.4 \times 10^{8} \mathrm{Pa} .$ What is the effective crosssectional area of the femur in a human and in a horse? (Note: since the center of the femur contains bone marrow, which has essentially no compressive strength, the effective cross- sectional area is about \(80 \%\) of the total cross-sectional area.)
A bob of mass \(m\) is suspended from a string of length \(L\) forming a pendulum. The period of this pendulum is \(2.0 \mathrm{s}\) If the pendulum bob is replaced with one of mass \(\frac{1}{3} m\) and the length of the pendulum is increased to \(2 L\), what is the period of oscillation?
The gravitational potential energy of a pendulum is \(U=m g y .\) (a) Taking \(y=0\) at the lowest point, show that \(y=L(1-\cos \theta),\) where \(\theta\) is the angle the string makes with the vertical. (b) If \(\theta\) is small, \((1-\cos \theta)=\frac{1}{2} \theta^{2}\) and \(\theta=x / L\) (Appendix A.7). Show that the potential energy can be written \(U=\frac{1}{2} k x^{2}\) and find the value of \(k\) (the equivalent of the spring constant for the pendulum).
See all solutions

Recommended explanations on Physics Textbooks

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