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3 years ago

6

A particle oscillates in simple harmonic motion, with amplitude A and period T. The particle starts from position x = A. What is

the magnitude of the particle's displacement a time 3T/2?
A. 3A
B. 6A
C. 2A
D. 0

1 answer:

frozen [14]3 years ago

8 0

Answer:

i guess it is D because,

y = A sin(wt)

w = $\frac{2\pi }{T}$ and t = 3T/ 2

now, Y = A sin ( $\frac{2\pi }{T} X \frac{3T}{2}$ )

so, Y = 0radian

Explanation:

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A person drives north 8 blocks, then turns west and drives 4 blocks. The driver then turns south and drives 8 blocks. How could

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A person would be driving 4 blocks west from the starting point to make the shorter distance while maintaining same displacement.

<u>Explanation:</u>

The measurement of an object’s position change from its point is called displacement. It is usually calculated from starting to the end points and represented by ‘delta s’. In the given scenarios, the person drove in the way that he finals the driving by 4 blocks away from the west.

Means, the persons drive to 8 blocks north and then to 8 blocks south get cancelled. Hence, to make the shorter distance with maintaining same displacement he would be driving 4 blocks west from the starting point.

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Which was a common goal of Spanish and British explorers in the Southeastern region of North America?

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Answer:

B

Explanation:

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A motorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. Calculate (a) the accelera

8090 [49]

Answer:

acceleration = -0.042 m/s²

velocity at beginning = 14.167 m/s

velocity at end = 5.7183 m/s

Explanation:

given data

distance d1 = 1 km

distance d2 = 2 km

time t1 = 80 s

time t2 = 120 s + 80s = 200 s

to find out

acceleration and velocity at beginning and end

solution

we apply here law of motion that is

d = vt + 1/2×at²

put value

1000 = v(80) + 1/2×a(80)² ........................1

and

2000 = v(200) + 1/2×a(200)² ........................2

so from equation 1 and 2 we get a and v

a = -0.042 m/s² and

v = 14.167 m/s

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V² = v² + 2ad

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Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. N

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Answer:

1. equilibrium

2. bottom

3. bottom

4. nowhere

5. bottom

6. top & bottom

7. equilibrium

8. equilibrium

1. No

2. Yes

Explanation:

According to the following equation of motion for SHM:

$x(t) = A\cos(\omega t + \phi)$

where A is the amplitude, ω is the angular frequency, and ∅ is the phase angle.

Furthermore, the velocity and acceleration functions are as follows:

$y(t) = -\omega A\sin(\omega t + \phi)\\a(t) = -\omega^2 A\cos(\omega t + \phi)$

1. The acceleration is zero at the equilibrium. At the equilibrium, the net force on the object is zero. And according to Newton's Second Law, if the net force is zero, then the acceleration is zero as well.

2. The forces on the object in a vertical spring are the weight of the object and the spring force.

F = mg - kx

Since mg is constant along the motion, then the net force is maximum at the amplitude. For the special case in this question, the mass is always below the rest length of the spring. So the net force is maximum at the lower amplitude, because x is greater in magnitude at the lower amplitude. According to Newton's Second Law, acceleration is proportional to the net force, hence the acceleration is at a maximum at the bottom.

3. As explained above, the magnitude of the net force is at a maximum at the lower amplitude, that is bottom.

4. The spring force is defined by Hooke's Law: F = -kx. Since the oscillation is small enough so that the mass is always below the rest length of the spring, then x is always greater than zero, hence nowhere in the motion will the spring force becomes zero.

5. As explained above, the force of gravity is constant and the spring force is proportional to the displacement, x. Therefore, the spring force is at a maximum at the lower amplitude, that is bottom.

6. The speed is zero when the mass is instantaneously at rest, that is the amplitude.

7. The net force on the mass is zero at the equilibrium.

8. The speed is at a maximum at the equilibrium.

1. We will use the equation of motions given above. For simplicity, let's take ∅ = 0. At half its amplitude:

$\frac{A}{2} = A\cos(\omega t)\\\frac{1}{2} = \cos(\omega t)\\\omega t = \pi / 3$

Then the velocity at that point is

$v(t) = -\omega A\sin(\pi /3) = -\omega A (0.866)$

The maximum speed is where the acceleration is equal to zero:

$0 = -\omega^2 A\cos(\omega t)\\\omega t = \pi / 2\\v_{max} = -\omega A\sin(\pi /2) = -\omega A$

Comparing the maximum velocity to the velocity at A/2 yields that it is not half the maximum velocity:

$-\omega A(0.866) \neq -\omega A$

2. The maximum acceleration is at the amplitude.

$A = A\cos(\omega t)\\\omega t = 2\pi\\a_{max} = -\omega^2 A\cos(2\pi) = -\omega^2 A$

And the acceleration at A/2 is

$\frac{A}{2} = A\cos(\omega t)\\\omega t = \pi / 3\\a(t) = -\omega^2 A\cos(\pi / 3) = -\omega^2 A (0.5)$

Comparing these two results yields that the acceleration at half the amplitude is half the maximum acceleration.

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