Physics question

Question

A point moves with simple harmonic motion whose period is 4 s. If it starts from rest

at a distance 4.0 cm from the centre of its path, find the time that elapses before it has

described 2 cm and the velocity it has then acquired. How long will the point take to reach

the centre of its path?

Answer 1

For harmonic motion at any given time displacement

x(t) = A sin ωt, (1)

where

A = 4 cm - amplitude,

ω = 2π/T - cyclic frequency,

T = 4s - period of oscillation,

t - time.

We assume that initial phase φ₀ is 0.

Because ω = 2π/4 = π/2, we can write

x(t) = A sin (π/2) t

We need to find time t when x = 2 cm or half of the amplitude A.

So we can write

A/2 = A sin (π/2) t

For sin we will get

sin (π/2)t = 1/2

(π/2) t = π/6

Hence time t = 1/3 s or 0.33 s

To find velocity we need to take a derivative from x(t) using equation (1)

Now we have

v (t) = dx/dt = A cos (π/2)t ·π/2

Plugin in for t = 1/3 s, we obtain for velocity

v(1/3) = 4 cos π/6 · π/2 = 4·√3/2 ·π/2 = √3π· = 1.73·3.14 = 5.43 m/s

At equilibrium position the point will be after 1 quarter of the period. Hence, t_1/4 = T/ 4 = 1s

It means it required time t₀ = t_1/4 - t to reach the center of the path.

Or

t₀ = 1s - 1/3 s = 2/3 s = 0.66 s

Answer 2

SETUP FOR THIS PROBLEM

Simple Harmonic Motion Equation:  A(t) = a*sin(ω*t)

Note: Watch the Units of measure in this problem!

A = cm as a function of time, t

a= cm

t= seconds

ω= radians / second

ω definition is: ω = (2π * 1/f )

where f is cycles/ sec and 2π is radians / cycle

f definition is 1 / p

where p is period in seconds

FROM THE PROBLEM INFORMATION

P = 4 sec which make f = 0.25 cycles/sec

a is 4cm because at peak amplitude the object is at rest & sin(ω*t) = π / 2 radians, ie

the peak of oscillation.

SOLUTION: Find T for A(t) = 2, the object moved from 4cm to 2 cm

A(t) = 2 cm = (4cm) * sin(2π * f * t )

Note: (2π * f * t ) is in radians

4cm / 2 cm = 1/2 = sin( 2 * 3.14 * 0.25 * t )

1/2 = sin( 1.57 * t)

ArcSin (1 /2 ) = 1.57 * t

the units are radians = radians

ArcSin (1 / 2) / 1.57 = t

Note: ArcSin asks " What angle has a sine of 1 /2?" it is 30° but what is it in

radians?

30° = 30° * 2π radians / 360° = 0.523 radians

ArcSin (1 / 2) / 1.57 = t

0.523 radians / (1.57 radians / second) = 0.333 seconds.

Answer #1: The object travels 2 cm in 0.333 second

SECOND SET UP:

 A(t) = a*sin(ω*t)

V(t) = d/dt (a*sin(ω*t) = d/dt (a * sin( 2π * 1/f *t) )

V(t) = a * cos ( 2π * 1/f * T ) * ( 2π * 1/f )

Where T is 0.25 sec + 0.333 seconds. The object is on the downward cycle of the

first half of the period and the velocity will negative! T= 0.583 seconds.

At t=0.333,

V(0.333) = 4 * cos ( 2 * 3.14 * 0.25 * 0.583) * (2 * 3.14 * 0.25 )

V(0.333) = 4 * cos (0.9153) * 1.57

For the student to confirm: 0.9153 radians = 142.47° & cos( 142.47°) = -0.455

Do the math!

ANSWER #2:

V(0.333) = 2.857 cm / sec

THIRD ANSWER:

The object is on the downward cycle of the first half of the period. In other wards on the downward slope of the second quarter cycle of the motion. That quarter cycle takes 1 second in total (Period = 4 sec). It has completed 0.333 second of that quarter cycle so 1.000 seconds - 0.333 seconds = 0.666 seconds remaining to reach the bottom of the second quarter cycle. That point is the midpoint of the full oscillation cycle and is at 0cm amplitude.