Question

(4%) Problem 9: A mass is connected to a spring and is allowed to move horizontally. The mass is at a position L when the spring is unstretched. The mass is then moved, stretching the spring, and released from rest. It then moves with simple harmonic motion. (a) At the instant that the mass passes through the position where the spring is unstretched, what can be said about its instantaneous acceleration? Grade Summary It is maximum. OIt is zero It is non-zero but not maximum Potential per attempt) Hint I give up 1% -ム33% Part (b) At the instant that the mass asses through the position where the spring is unstretched, what can be said about its instantaneous ーム33% Part (c) At the instant that the spring is compressed the most, what can be said about the magnitude of its instantaneous velocity?

Answer 1

Answer:

a) Acceleration is zero , c)   Speed ​​is cero

Explanation:

a) the equation that governs the simple harmonic motion is

         x = A cos (wt +φφ)

Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition

Body acceleration is

         a = d²x / dt²

Let's look for the derivatives

         dx / dt = - A w sin (wt + φ)

         a = d²x / dt² = - A w² cos (wt + φ)

In the instant when it is not stretched x = 0

As the spring is released at maximum elongation, φ = 0

            0 = A cos wt

            Cos wt = 0         wt = π / 2

Acceleration is valid for this angle

           a = -A w² cos π/2 = 0

Acceleration is zero

b)

c) When the spring is compressed x = A

Speed ​​is

             v = dx / dt

             v = - A w sin wt

We look for time

            A = A cos wt

            cos wt = 1         wt = 0, π

For this time the speedy vouchers

            v = -A w sin 0 = 0

Speed ​​is cero