All categories

4 years ago

For a spring oscillating in simple harmonic motion, at what point will the velocity of

Physics

1 answer:

Crazy boy [7]4 years ago

6 0

Answer:

equilibrium position.

Explanation:

In simple harmonic motion , velocity v(t) is given by,

v(t) = -ω A sin(ωt + φ)

where

ω = angular velocity of the corresponding circular motion

A = amplitude

t = time

φ = the initial angle of the corresponding circular motion when the motion begin.

v (t) get maximized when sin value is maximized , i.e. sin $\alpha$ =1

The particle has maximum speed when it passes through the equilibrium position.

You might be interested in

A 300 g wooden block on a smooth, level surface is firmly attached to a very light horizontal spring with a spring constant of 2

Tom [10]

the solution for the oscillatory movement allows to find the result for the amplitude of the initial displacement is:

The range of motion is: A = 4.44 cm

<h3>Oscillatory movement.</h3>

The oscillatory periodic motion of a system occurs when there is a recovered force, in the special case that this force is proportional to the displacement is called simple harmonic motion, which is described by the expression.

x = A cos (wt + Ф)

w² = k/m

where x is the displacement, A the amplitude, w the angular velocity, t the time, k the spring constant, m the mass, and Ф a phase constant determined by the initial conditions.

Let's find the angular velocity/

w= $\sqrt{ \frac{200}{0.300} }$

w = 25.8 rad/s

Let's look for the constant Ф, as the system is released from rest its initial velocity is zero, for zero time. The definition of speed is:

v= $\frac{dx}{dt}$

v= - A w sin (wt +Ф)

0 = -A w sin Ф

Ф= 0

They indicate that at a given instant of the time the velocity is v= 50.0 cm/s and it is in a position x= 4.00 cm, let us write the equations for this time

Position.

4.00 = A cos 25.8t

Speed.

50.0 = - At 25.8 sin 25.8t

To solve the system, ;et's square and add.

Cos² 25.8t = $\frac{16}{A^2}$

sin² 25.8t = $\frac{3.756}{A^2 }$

1 = $\frac{1}{A^2} \ (16 + 3.756)$

A = $\sqrt{19.756}$

A= 4.44 cm

In conclusion using the solution for the oscillatory movement we can find the result for the amplitude of the initial displacement is:

The range of motion is: A = 444 cm

Learn more about oscillatory motion here: brainly.com/question/14311816

4 0

2 years ago

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum

asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0

3 years ago

Can somebody please help me!

Serga [27]

Answer:

graph it like thisdo the number it has so do 14

Explanation:

so do 14 and you have it

6 0

3 years ago

A 12 kg block is accelerating at the rate of 9.0 m/s? while being acted on by two forces.

Aliun [14]

Answer:

See below

Explanation:

F = ma

F = 12 * 9 = 108 N

108 N needed <u> add 30 N more east </u>

8 0

1 year ago

An Olympic discus thrower (~100 kg) launches the 2.0 kg discus by spinning rapidly (~4 times per second) with arm outstretched (

vladimir1956 [14]

Answer:

F = 1263.03 N

Explanation:s

given,

mass of the disk thrower = 100 Kg

mass of the disk = 2 Kg

angular speed of the disk = 4 rev/s

arm outstretched = 1 m

centripetal force of the disk in the circular path

F = m ω² r

ω = 4 x 2 x π

ω = 25.13 rad/s

F = m ω² r

F = 2 x 25.13² x 1

F = 1263.03 N

hence, centripetal force equal to the F = 1263.03 N

6 0

3 years ago