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Sphinxa [80]
3 years ago
11

A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its maximum displacement from

equilibrium, what is its instantaneous velocity?A. At maximum displacement, its instantaneous velocity is less than its maximum but not zero.B. Instantaneous velocity cannot be determined without additional information.C. At maximum displacement, its instantaneous velocity is zero.D. At maximum displacement, its instantaneous velocity is also at maximum.
Physics
1 answer:
Mrac [35]3 years ago
4 0

Answer:

C) True. At maximum displacement, its instantaneous velocity is zero.

Explanation:

The simple harmonic movement is given by

        x = A cos wt

Speed

        v = - A w sin wt

At the point of maximum displacement x = A

       A = A cos wt

      cos wt = 1

      wt = 0

We replace the speed

       v = -Aw sin 0 = A w

Speed ​​is maximum

Let's review the claims

A) False. Speed ​​is zero

B) False. It can be determined

C) True. Agree with our result

D) False. When one is maximum the other is minimum

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A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
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Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

E_{A}=E_{B}

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

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E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

E_{B}=m*g*h+\frac{1}{2}*m*v^{2}

Therefore we will have the following equation:

(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]

The kinetic energy can be easily calculated by means of the kinetic energy equation.

KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]

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