Figure 15-24shows the x(t) curves for three experiments involving a particular spring–box system oscillating in SHM. Rank the curves according to (a) the system’s angular frequency, (b) the spring’s potential energy at time t=0, (c) the box’s kinetic energy att=0, (d) the box’s speed att=0, and (e) the box’s maximum kinetic energy, greatest first.

Short Answer

Expert verified

a) Ranking of curves according to system’s angular frequency isω1=ω2=ω3

b) Ranking of curves according to springs potential energy at t = 0 is U3>U2=U1.

c) Ranking of curves according to box’s kinetic energy at t = 0 is K.E1>K.E2>K.E3.

d) Ranking of curves according to box’s speed at t = 0 is v1>v2>v3.

e) Ranking of curves according to box’s maximum kinetic energy is role="math" localid="1657260771388" K.Emax1>K.Emax3>K.Emax2.

Step by step solution

01

The given data 

The graph of position versus time for SHM of spring-box system is given.

02

Understanding the concept of SHM of a particle

Using the formula for angular frequency which is related to spring constant and mass, we can rank the curves. From the spring’s potential energy formula, we can rank the curves, and from the formula of kinetic energy, we can rank the curves for kinetic energy. We can also rank the speed of the box. For ranking maximum kinetic energy, we consider the amplitude of the curves.

Formulae:

The angular frequency of a body in SHM,ω=km (i)

The potential energy of a spring system,U=12kx2 (ii)

The kinetic energy of a body in SHM, K.E=12mv2 (iii)

03

Calculation of the ranking of curves according to the system’s angular frequency

a)

As we know the mass of the box and the spring constant is the same for the same oscillatory system while doing the experiment, therefore, the angular frequency will be the same for these curves considering equation (i).

Hence, ranking of angular frequencies is ω1=ω2=ω3.

04

Calculation of the ranking of curves according to potential energy of the spring

b)

From equation (ii), we can see that the potential energy of the spring depends on the displacement.

Again, in the graph at t = 0 we get the displacement of curve 3 is greater than the other two. The displacements of curve 1 and 2 are same.

Hence, the ranking of potential energies is U3>U2=U1.

05

Calculation of the ranking of curves according to box’s kinetic energy

c)

Slope of the curves gives the velocity, v=xt

From the graph, the rank of the velocities at t = 0 can be given as:v1>v2>v3.

Therefore, the ranking of kinetic energies using equation (iii) is K.E1>K.E2>K.E3.

06

Calculation of the ranking of curves according to box’s speed

d)

From part (c), we have seen that ranking of velocities as:v1>v2>v3 , and since all have same direction the ranking of speed is same as the velocities at t = 0.

Therefore, the ranking of speed is v1>v2>v3.

07

Calculation of the ranking of curves according to maximum kinetic energy of the box

e)

The kinetic energy is proportional to the amplitude of the curve, so from the graph we get,

The amplitude of the curves as:1>3>2 .

The ranking of maximum kinetic energies is K.Emax1>K.Emax3>K.Emax2.

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 2.00 kgblock hangs from a spring. A 300 kgbody hung below the block stretches the spring 2.00 cmfarther.

  1. What is the spring constant?
  2. If the 300 kgbody is removed and the block is set into oscillation, find the period of the motion.

Figure 15-38 gives the one-dimensional potential energy well for a 2.0 Kgparticle (the function U ( x )has the formbx2and the vertical axis scale is set byUs=2.0J).

  1. If the particle passes through the equilibrium position with a velocity of, 85 cm / s will it be turned back before it reaches x = 15 cm?
  2. If yes, at what position, and if no, what is the speed of the particle at x = 15cm?

In Figure, a block weighing 14.0 N, which can slide without friction on

an incline at angle40.0, is connected to the top of the incline by a massless

spring of unstretched length 0.450 mand spring constant 120 N/m.

a) How far from the top of the incline is the block’s equilibrium point?

b) If the block is pulled slightly down the incline and released, what is the period

of the resulting oscillations?

In Figure 15-37, two blocks(m=1.8kgandM=10kg)(and) and a spring (k=200 N/m) are arranged on a horizontal, frictionless surface. The coefficient of static friction between the two blocks is 0.40.What amplitude of simple harmonic motion of the spring–blocks system puts the smaller block on the verge of slipping over the larger block?

A 3.0kg particle is in simple harmonic motion in one dimension and moves according to the equation x=(5.0 m)cos [(ττ/3rad/s)t-ττ/4rad],with t in seconds. (a) At what value of x is the potential energy of the particle equal to half the total energy? (b) How long does the particle take to move to this position x from the equilibrium position?

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