Physics, 21.10.2020 03:01 RealSavage4Life

An object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in cm and t in seconds. Give decimal answers below. (a) How many complete back-and-forth motions (from the origin to the right, back to the origin, to the left and finally back to the origin) does the object make in one second? (b) What is t the first time that the object is at its farthest right? (c) At the time found in part (b), what is the object's velocity? (d) At the time found in part (b), what is the object's acceleration?

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Answer from: diego4325

The type of motion the object undergoes that consist of a back-and-forth

movement that is repetitive is a simple harmonic motion.

The correct responses are;

(a) Number of complete back-and-forth motions per second is 10(b) The first time is after 0.025 seconds(c) Velocity is 0(d) Acceleration is approximately -11843.53 m/s²

Reasons:

The equation that represents the motion of the object is x(t) = 3·sin(20·π·t)

Where;

t ≥ 0

x(t) is in cm, and t is in seconds.

(a) The general sine function is y = A·sin(B·(t - C) + D

$\mathrm{The \ period \ P} = \dfrac{2 \cdot \pi}{B}$

By comparison, we have;

B = 20·π

Therefore;

$P = \dfrac{2 \cdot \pi}{20 \cdot \pi} = \dfrac{1}{10} = 0.1$

The period, which is the time to complete one cycle = 0.1 seconds

The number of cycle completed per second, is the frequency, f

$f = \dfrac{1}{P} = \dfrac{1}{0.1} = 10$

f = 10 Hz = 10 cycles

The number of cycle completed per second is f = 10 cycles

(b) When the object is at the farthest right, we have;

sin(20·π·t) = Maximum = 1

$\mathbf{sin\left(\dfrac{\pi}{2}\right)} = 1$

Therefore,

$20 \cdot \pi \cdot t = \dfrac{\pi}{2}$

$t = \dfrac{\pi}{20 \cdot \pi \cdot 2} = \dfrac{1}{40} = 0.025$

The first time it is at the farthest right is t = 0.025 seconds after start

Therefore;

$v = \dfrac{dx(t)}{dt} = \dfrac{dx(t)}{dt} \left(3 \cdot sin \left(20 \cdot \pi \cdot t \right) = 60\cdot \pi \cdot cos \left(20 \cdot \pi \cdot t \right)$

At t = 0.25 seconds, we have;

$v = \dfrac{dx(t)}{dt} = 60\cdot \pi \cdot cos \left(20 \times \pi \times 0.025 \right) = 0$

The velocity of the object at the t = 0.025 seconds is v(t) = 0

(d) Acceleration, a, is given by $a = \dfrac{dv(t)}{dt}$

Therefore;

$a = \dfrac{dv(t)}{dt} = \dfrac{d}{dt} \left( 60\cdot \pi \cdot cos \left(20 \times \pi \times t\right) \right) =-1200 \cdot \pi ^2 \cdot sin(20 \cdot t \cdot \pi)$

a = -1200·π²·sin(20·t·π)

∴ a = -1200×π²×sin(20×0.025×π)

sin(20×0.025×π) = 1

∴ a = -1200×π²×1 ≈ -11843.53

At the time found in part b the acceleration of the object, a ≈ -11843.53 m/s²

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Answer from: mckinleesmomp6qj1e

a.) 10Hz

b.) 0.1 s

c.) 187.4 m/s

d.) -412.6 m/s^2

Explanation:

Given that an object is moving back and forth on the x-axis according to the equation x(t) = 3sin(20πt), t> 0, where x(t) is measured in cm and t in seconds. Give decimal answers below.

(a) How many complete back-and-forth motions (from the origin to the right, back to the origin, to the left and finally back to the origin) does the object make in one second?

from the equation given, the angular speed w = 20π

but w = 2πf

where f = frequency.

substitute w for 20π

20π = 2πf

f = 20π/2π

f = 10 Hz

(b) What is t the first time that the object is at its farthest right?

since F = 1/T

T = 1 / f

T = 1/10

T = 0.1 s

Therefore, the t of first time that the object is at its farthest right is 0.1 s

The velocity can be found by differentiating the equation;

x(t) = 3sin(20πt)

dx/dt = 60πcos(20πt)

where dx/dt = velocity V

V = 60πcos(20π * 0.1)

V = 187.4 m/s

(d) At the time found in part (b), what is the object's acceleration?

to get the acceleration, differentiate equation V = 60πcos(20πt)

dv/dt = -1200πSin(20πt)

dv/dt = acceleration a

a = -1200πSin(20πt)

substitute t into the equation

a = -1200πSin(20π * 0.1)

a = - 412.6 m/s^2

Answer from: Quest

i. like it i'm gonna edit it tomorrow

Answer from: Quest

i will answer your question if you answer mine

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