/*! 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} Q. 46 A 200 g block hangs from a spr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 200gblock hangs from a spring with spring constant 10N/m. At the block islocalid="1650020763199" 20cmbelow the equilibrium point and moving upward with a speed oflocalid="1650020788839" 100cm/s. What are the block's

a. Oscillation frequency?

b. Distance from equilibrium when the speed is 50cm/s?

c. Distance from equilibrium at t=1.0s?

Short Answer

Expert verified

a. Oscillation frequency is 1.1Hz.

b. Distance is 23cm.

c. Distance is-4.1cm.

Step by step solution

01

Calculation for frequency of oscillation (part a)

a.

Given info:

The mass is,

m=(200g)×1kg1000g

=0.2kg

The spring constant is10N/m.

Oscillation frequency is,

f=12Ï€km

f=12×3.1410N/m0.2kg

f=1.1Hz

02

Calculation for angular frequency and function of time

Angular frequency is,

Ó¬=Km

Ó¬=10N/m0.2kg

Ó¬=52s-1

Function of time is,

y(t)=Acos(Ó¬t+Ï•)

For time is zero means,

y(0)=Acos(Ï•)

=-20cm

03

Calculation of speed

Speed is,

vy(t)=dy(t)dt

=ddt[Acos(Ó¬t+Ï•)]

vy(t)=-Ó¬A[sin(Ó¬t+Ï•)]

Where time is zero,

vy(0)=-Ó¬A[²õ¾±²ÔÏ•)]=100cm/s

From that,

vy(0)y(0)=-Ó¬tan(Ï•)=100cm/s-20cm

-Ó¬tan(Ï•)=-5s-1

=-2.526rad

04

Calculation for time

Amplitude is ,

A=-20cmcos(-2.526rad)

=24.5cm

Time calculated as,

vy(t)=-Ó¬Asin(Ó¬t+Ï•)

Ó¬t+Ï•=sin-1-50cm/sÓ¬A

t=sin-1-50cm/sӬA-ϕӬ

t=sin-1-50cm/s52s-1(24.5cm)-(-2.526rad)52s-1

=0.3158s

05

Calculation of distance (part b)

b.

When speed is 50cm/s,

x(0.3158s)=(24.5cm)cos52s-1(0.3158s)-2.526rad

=23cm

06

Calculation for distance (part c)

c.

Where t=1.0s,

x(1.0s)=Acos(Ó¬t+Ï•)

=(24.5cm)cos52s-1(1.0s)+2.526rad

=-4.1cm

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 91Ó°ÊÓ!

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

It is said that Galileo discovered a basic principle of the pendulum—

that the period is independent of the amplitude—by using

his pulse to time the period of swinging lamps in the cathedral

as they swayed in the breeze. Suppose that one oscillation of a

swinging lamp takes5.5s.

a. How long is the lamp chain?

b. What maximum speed does the lamp have if its maximum

angle from vertical is 3.0∘?

FIGURE Q15.9 shows the potential-energy diagram and the total energy line of a particle oscillating on a spring.

a. What is the spring's equilibrium length?

b. Where are the turning points of the motion? Explain.

c. What is the particle's maximum kinetic energy?

d. What will be the turning points if the particle's total energy is doubled?

What is the difference between the driving frequency and the natural frequency of an oscillator?

The analysis of a simple pendulum assumed that the mass was a particle, with no size. A realistic pendulum is a small, uniform sphere of mass Mand radius Rat the end of a massless string, withLbeing the distance from the pivot to the center of the sphere.

a. Find an expression for the period of this pendulum.

b. SupposeM=25g,R=1.0cm,andL=1.0m, typical values for a real pendulum. What is the role="math" localid="1650084928460" Treal/Tsimpleratio, whererole="math" localid="1650084943350" Trealis your expression from part a androle="math" localid="1650084959286" Tsimple is the expression derived in this chapter?

Astronauts on the first trip to Mars take along a pendulum that has a period on earth of 1.50s. The period on Mars turns out to be 2.45s. What is the free-fall acceleration on Mars?

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.