A 350gmass on a 45-cm-long string is released at an angle of 4.5°from vertical. It has a damping constant of 0.010kg/s. After 25s,

(a) how many oscillations has it completed and

(b) how much energy has been lost?

Short Answer

Expert verified

(a) Number of oscillationn=18.6

(b)Elost=2.4×103Jis the energy lost.

Step by step solution

01

Concept 

Formula for displacement oscillation is

Equation1

x(t)=Aebt/2mcos(ωt+ϕ)

Damped oscillation has a equation of

Equation2

ω=gLb24m2

Dampled oscillation is varied in time

Equation 3

E(t)=E0ebt/m

Terms of frequency in a wave

Equation 4

ω=2πf

Formula for gravitation potential energy

Equation 5

Ug=mgy

02

Find the number of oscillations (part a)

Equation (2) yields the frequency content of the decelerated oscillator:

ω=gLb24m2

Fill up the blanks with numerical values from the given data:

ω=9.80m/s20.45m(0.010kg/m)24(0.350kg)2

=4.666644801s1

Equation (4) yields the phase of the damped oscillator:

T=2πω

=2π4.666644801s1

=1.346s

The oscillator obtained the required number of oscillations in 25s:

n=tT

=25s1.346s

=18.6oscillations

03

Find the lost energy (part b)

Allow the string's gravitational potential energy to be 0 at its lowest point. Equation (5) shows that the string possesses a gravitational potential energy at first:

E0=Ug=mgh

The height is calculated using θ, the angle, and L, the string length. As a result, h-L(1-cosθ)

E0=mgL(1cosθ)

Equation (3) expresses the string's energy as a function of time:

E(t)=E0ebt/m

The heat absorbed by that of the string after 25is equal to the string's original energy minus the stage's energy at time 25.t=25sSo

Elost=E0E(t)

=E0E0ebt/m=E01ebt/m

=mgL(1cosθ)1ebt/m

Numericals values are substituted

role="math" localid="1649966946821" Elost=(0.350kg)9.80m/s2(0.45m)1cos4.5,1exp(0.010kg/s)(25s)0.350kg

=2.4×103J

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