Question

You are to complete Figure 15-22a so that it is a plot of velocity v versus time t for the spring–block oscillator that is shown in Figure 15-22b for t=0. (a) In Fig.15-22a, at which lettered point or in what region between the points should the (vertical) v 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 velocity is given by$\mathit{v}\mathbf{=}\mathbf{-}{\mathit{v}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{\omega }\mathbf{t}\mathbf{+}\mathbf{\varphi }\mathbf{)}$, 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$\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{rad}$).

Short answer

a) Between points D and E v-axis intersects the t-axis.

b) The value of $\varphi$if the velocity is given $v=v-v_m\sin \left(\omega t+\varphi \right)$is between$\frac{3\mathrm{\pi }}{2}\mathrm{and}2\mathrm{\pi }$ .

Step-by-step solution

The given data

a) The graph for v versus t of SHM and figure for block-spring system.

b) The velocity of the block is given as,$v=-v_m\sin \left(\omega +\varphi \right)$ .

Understanding the concept of SHM of a particle

We can use the concept of SHM. From the given graph, we can determine the point at which the v-axis intersects with the time axis from the position of the block and whether the velocity is decreasing or increasing as the block moves away from the mean position. Also, from the velocity equation, we can find the range of the phase angle.

Formula:

The velocity equation of a body in SHM, $v=-v_m\sin \left(\omega +\varphi \right)$ (i)

Calculation of the points at which v-axis intersects t-axis

a)

From the figure, we can see that the block is at $\frac{X_m}{2}$ position when=0, and it is moving along positive direction.

Block is moving away from the mean position so the velocity is decreasing. So from both the figures, we can determine thatthe block is between point D and E.

Hence, the velocity axis will intersect in region D and E.

Calculation of the phase angle from velocity equation

b)

Att=0,we get the velocity equation using equation (i) as follows:

$v=v-\sin \left(\varphi \right)$

From this equation, we can determine that at t=0 velocity is positive and decreasing. So, if velocity is positive, it means that the value of$\sin \left(\varphi \right)$must be negative. We know that$\sin \left(\varphi \right)$negative value is in III and IV quadrant. From the figure, the block is between point D and E.

Hence, the values of$\varphi$ is between$\frac{3}{2}\mathrm{and}2\mathrm{\pi }$.