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lisov135 [29]
4 years ago
10

If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat

or change?
Physics
1 answer:
Maksim231197 [3]4 years ago
6 0

Answer:

If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2

Explanation:

We know that in a simple harmonic oscillator the maximum speed is given by

    v_{max} = Aw

  Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to w\\ of the oscillation .

  Since  w = 2 \pi f

      v_{max}^{|}/v_{max} = 2f/f = 2

  Where v_{max}^{|} is the maximum speed when frequency is doubled .

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When a force is exerted on an object, work is done only if the object
Kazeer [188]

Is moving, i'm almost positive

4 0
3 years ago
a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por q
Tatiana [17]

Answer:

a) the correct answer is 1 , b) x=0   F=0, a=0

x= 0.050    F= -7.5 N,  a= -15 m/s²

x= 0.150     F= 22.5 N,  a=- 45 m/s²

Explanation:

a) In a mass - spring system the force is given by the Hooke force,

          F = - k x

Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position

the correct answer is 1

b) we assume that the given values ​​are from the equilibrium position of the spring.

Let's calculate the force

x = 0

      F = 0

x = 0.050

      F = - 150 0.050

      F = - 7.50 N

x = 0.150

      F = - 150 0.150

      F = - 22.5 N

let's use Newton's second law to find the acceleration

      F = m a

      a = F / m

x = 0 m

      a = 0

x = 0.050 m

      a = -7.50 / 0.50

      a = - 15 m / s²

x = 0.150 m

      a = - 22.5 / 0.50

      a = - 45 m/s²

TRASLATE

a) En un sistema masa – resorte  la fuerza es dada por la fuerza de Hoke,  

          F= - k x

analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio

la respuesta correcta es  1

b)suponemos que los valores dados son desde la posición de equilibrio del resorte.

Calculemos la fuerza  

x=0  

              F= 0

x=0.050  

              F = - 150 0.050

              F= - 7.50 N

x= 0.150  

                F= - 150 0.150

                F= - 22.5 N

usemos la segunda ley de Newton para encontrar la aceleración

          F = m a

          a = F/m

x =0  m

        a = 0

x= 0.050 m

         a = -7.50/ 0.50

          a =- 15 m/s²

x= 0.150 m

          a= - 22.5 / 0.50

          a= - 45 m/s²

7 0
4 years ago
At an accident scene on a level road, investigators measure a car's skid mark to be 88 m long. It was a rainy day and the coeffi
german

Answer:

The speed was 26.91m/s (96.9 km/h)

Explanation:

Here you have to consider that at the beginning you have an amount of kinetic energy (K) that is dissipated because of the work done by friction forces (T). Since the car is stopped after the accident, all the energy has been dissipated. Thus, K=T.

The definition of the kinetic energy is K=1/2*m*v^2.

The work done by the friction forces is: T=f*(m*g)*d. Where <em>f</em> is the friction coefficient, <em>g</em> is the gravity acceleration, <em>m i</em>s the mass of the car and <em>d</em> is the skid marks longitude. Therefore,

1.2*m*v^2=f*(m*g)*d

Since <em>m</em> is in both sides it can be cancelled so it is not necessary to considered.

Then, the speed is determined by the following equation:

v=\sqrt{2*f*g*d} =\sqrt{2*0.42*9.8m/s^2*88 m}=26.91 m/s

4 0
3 years ago
Definition: The energy transferred by a force to a moving object.
Marysya12 [62]

Answer:

Force applied

Explanation:

An object will remain at its state of rest unless a non zero for act on it

6 0
3 years ago
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8
hammer [34]
Missing figure and missing details can be found here:
<span>http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....
</span>
Solution:
(a) The work done by the spring is given by
W= \frac{1}{2} k (\Delta x)^2 &#10;
where k is the elastic constant of the spring and \Delta x is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
W_W = -F_{//} (x_2 -x_1)
where  the negative sign is given by the fact that F_{//} points in the opposite direction of the displacement of the cart, and where
F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N
therefore, the work done by the weight is
W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J

8 0
3 years ago
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