Question

current attempt in progress
the drawing shows a graph of displacement x versus time t for simple harmonic motion of an object on a horizontal spring. which one of the following answers correctly gives the magnitude v of the velocity and the magnitude a of the acceleration at points a and b in the graph?

$v_a = 0 m/s, a_a = maximum, v_b = maximum, a_b = 0 m/s^2$
$v_a = maximum, a_a = maximum, v_b = 0 m/s, a_b = maximum$
$v_a = maximum, a_a = 0 m/s^2, v_b = 0 m/s, a_b = maximum$
$v_a = 0 m/s, a_a = 0 m/s^2, v_b = maximum, a_b = maximum$
$v_a = maximum, a_a = maximum, v_b = 0 m/s, a_b = 0 m/s^2$

Explanation:

Step1: Recall the properties of simple - harmonic motion

In simple - harmonic motion, the velocity \(v\) and acceleration \(a\) are related to the displacement \(x\). The velocity \(v\) is given by \(v =-\omega\sqrt{A^{2}-x^{2}}\) (where \(A\) is the amplitude and \(\omega\) is the angular frequency) and the acceleration \(a=-\omega^{2}x\).

Step2: Analyze point \(A\)

At point \(A\), \(x = 0\).

• For velocity: When \(x = 0\), \(v=\omega A\) (maximum velocity).
• For acceleration: When \(x = 0\), \(a =-\omega^{2}x=0\).

Step3: Analyze point \(B\)

At point \(B\), \(x = A\) (amplitude).

• For velocity: When \(x = A\), \(v =-\omega\sqrt{A^{2}-A^{2}}=0\).
• For acceleration: When \(x = A\), \(a=-\omega^{2}A\) (maximum magnitude of acceleration since \(|a|=\omega^{2}A\)).

Answer:

\(v_{A}=\text{maximum},a_{A}=0\space m/s^{2},v_{B}=0\space m/s,a_{B}=\text{maximum}\)