91Ó°ÊÓ

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

b) The velocity of the block is given as,$v=-v_m\sin \left(\mathrm{Ó\neg }+\mathrm{Ï•}\right)$ .

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, $v=-v_m\sin \left(\mathrm{Ó\neg }+\mathrm{Ï•}\right)$ (i)

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.

b)

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

$v=v-\sin \left(\mathrm{Ï•}\right)$

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$\sin \left(\mathrm{Ï•}\right)$must be negative. We know that$\sin \left(\mathrm{Ï•}\right)$negative value is in III and IV quadrant. From the figure, the block is between point D and E.

Hence, the values of$\mathrm{Ï•}$ is between$\frac{3}{2}\mathrm{and}2\mathrm{Ï€}$.