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serious [3.7K]
2 years ago
10

The position of an object in simple harmonic motion is defined by the function y = (0.50 m) sin (πt/2). Determine the maximum sp

eed of the object.
Physics
1 answer:
Gnoma [55]2 years ago
5 0

The maximum speed of the object under simple harmonic motion is 0.786 m/s.

The given parameters:

  • Position of the particle, y = 0.5m sin(πt/2)

<h3>Wave equation for simple harmonic motion;</h3>

y = A sin(ωt + Ф)

where;

  • A is the amplitude = 0.5 m
  • ω is the angular speed = π/2

The maximum speed of the object is calculated as follows;

V_{max} = A \omega\\\\V_{max} = 0.5 \times \frac{\pi}{2} = \frac{\pi}{4} \ m/s  = 0.786 \ m/s

Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.

Learn more about simple harmonic motion here: brainly.com/question/17315536

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Suppose a rock is dropped off a cliff with an initial speed of 0m/s. What is the rocks speed after 5 secounds, in m/s, if it enc
Tasya [4]

Answer:

The rock's speed after 5 seconds is 98 m/s.

Explanation:

A rock is dropped off a cliff.

It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².

Speed after 5 seconds = V

We know that acceleration = average speed/time

In our case,

g = ((0+V)/2)/5

9.8*5 = V/2

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V = 98 m/s

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3 years ago
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La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit
katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ )  m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

         v = dr / dt

the function of position

         r = 2 cos πt  i^  + 3 sin πt  j^

let us note that it is a movement in two dimensions

let's perform the derivative

          v = -2π sin πt  i^  + 3π cos πt  j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

          v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

           v = (-4.44 i^ + 6.66 j^ )  m/s

To calculate the mean acceleration we use the expression

           a = (v_{f} - v_{o}) / Δt

 

indicates that the time is the first 3 s

       

we look for the initial velocity t = 0 s

           v₀ = 0 i ^ + 3π j ^

         

we look for the fine velocity, t = 3 s

          v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

          v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

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6 0
4 years ago
An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec
jolli1 [7]

Answer:

v_o=39\ m/s\\t_m=4\ s

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

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(a)

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(b)

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\displaystyle t_m=\frac{v_o}{g}

\displaystyle t_m=\frac{39}{9.8}

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3 years ago
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