You are to complete Fig 15-23aso that it is a plot of acceleration a versus time t for the spring–block oscillator that is shown in Fig 15-23b for t=0 . (a) In Fig.15-23a, at which lettered point or in what region between the points should the (vertical) a axis intersect the t axis? (For example, should it intersect at point A, or maybe in the region between points A and B?) (b) If the block’s acceleration is given by$\mathit{a}\mathbf{=}\mathbf{-}{\mathit{a}}_{\mathbf{m}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{\omega }\mathbf{t}\mathbf{+}\mathbf{\varphi }\mathbf{)}$what is the value of$\mathit{\varphi }$? Make it positive, and if you cannot specify the value (such as+$\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{rad}$), then give a range of values (such as between 0 and$\mathbf{\pi }\mathbf{/}\mathbf{2}$).

• The graph acceleration versus time and the figure for the block-spring system.
• The acceleration is given as$a=-a_m\cos \left(\omega t+\varphi \right)$ .

We can use the concept of SHM. From the given graph, we can determine the point at which the acceleration axis intersects the time axis from the position of the block and the acceleration whether is decreasing or increasing from the motion of the block. Using the given acceleration equation we can find the angle.

Formula:

The displacement equation of a particle in motion,$a=-a_m\cos \left(\omega t+\varphi \right)$ (i)

where,$a_m$ is the maximum displacement, $\omega$is the angular velocity, t is time and$\varphi$ is the phase difference.

From the given graph and the block spring diagram, we can see that the block is moving towards$-x_m$ from mean position. Att=0, the block is at point C. It moves from 0 to$-x_m$means its acceleration is changing from 0 to$-a_m$It implies that acceleration is decreasing. So it is moving towards point C.

Therefore, between B and C the acceleration axis intersects time axis.

From the graph att=0 the acceleration is positive; we get the acceleration equation using equation (i) as follows:

$a=-a_m\cos \left(\varphi \right)$

As the acceleration is positive the angle must be negative, so the cosine function is negative in quadrant II.

Hence, from the graph the value of$\varphi$ is in the range is $\frac{\mathrm{\pi }}{2}\left(rad\right)to\mathrm{\pi }\left(\mathrm{rad}\right)$.