The instantaneous acceleration is zero when the mass passes through the position where the spring is unstretched.
The equation that governs simple harmonic motion is as follows:
x = A cos (wt +φφ)
Where A is the movement's amplitude, w is the angular velocity, and the initial phase φ is determined by the initial condition.
The body accelerates.
a = d²x / dt²
Look for the derivatives.
dx / dt = - A w sin (wt + φ)
a = d²x / dt² = - A w² cos (wt + φ)
When it is not stretched, x equals 0.
When the spring is at its maximum elongation, φ = 0
0 = A cos wt
Cos wt = 0 wt = π / 2
This angle accepts acceleration.
a = -A w² cos π/2 = 0
Therefore acceleration is zero
x = A when the spring is compressed
Then the speed is
v = dx / dt
v = - A w sin wt
We are looking for time.
A = A cos wt
cos wt = 1 wt = 0, π
For this period, the quick vouchers
v = -A w sin 0 = 0
The speed is zero.
The complete question:
" 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. Part (b) At the instant that the mass asses through the position where the spring is unstretched, what can be said about its instantaneouPart (c) At the instant that the spring is compressed the most, what can be said about the magnitude of its instantaneous velocity?"
To learn more about acceleration refer to :
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