91Ó°ÊÓ

Velocity at any point x is given as,

${\mathit{V}}_{\mathbf{x}}\mathbf{Â\pm }\sqrt{\frac{\mathbf{K}}{\mathbf{m}}}\sqrt{{\mathbf{A}}^{\mathbf{2}}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}$ ……….. (1)

Where K is the spring constant, A is the amplitude of the SHM, and x denotes position of object.

Maximum speed occurs when it is passes through x=0,

$V_{max}=Â\pm \sqrt{\frac{K}{m}}A$ ……….. (2)

We know that angular frequency is given as,

role="math" localid="1668093865478" $\mathrm{Ó\neg }=\sqrt{\frac{K}{\mathrm{m}}}$ ……….. (3)

Hence,

$V_{max}=Â\pm \mathrm{Ó\neg }A$ ……….. (4)

We know that is given as,

$\mathbf{Ó\neg }\mathbf{=}\frac{\mathbf{2}\mathbf{Î }}{\mathbf{T}}$ ……….. (5)

Hence, from equation (4) and (5)

${\mathrm{v}}_{\max }=\frac{2\mathrm{Π}}{\mathrm{T}}\mathrm{A}$ ……….. (6)

From graph the maximum value of displacement A=10 cm =$10×{10}^{-2}\mathrm{m}$.

From graph the value of time period T=16 sec.

a)

From the equation derived above,

$\begin{array}{r}{\mathrm{v}}_{max}=\frac{2\mathrm{Π}}{\mathrm{T}}\mathrm{A}\\ =\frac{2\mathrm{Π}}{16\mathrm{s}}×\left(10×{10}^{−2}\mathrm{m}\right)\\ =0.039\mathrm{m}/\mathrm{s}\\ =3.9\mathrm{cm}/\mathrm{s}\end{array}$

b)

Maximum acceleration is given by,

$\begin{array}{r}\left|a_{max}\right|={\mathrm{Ó\neg }}^2\mathrm{A}\\ ={\left(\frac{2\mathrm{Î }}{T}\right)}^2\mathrm{A}\\ ={\left(\frac{2\mathrm{Î }}{16\mathrm{s}}\right)}^2×\left(10×{10}^{−2}\mathrm{m}\right)\\ =0.0154\mathrm{m}/{\mathrm{s}}^2\\ =1.54\mathrm{cm}/{\mathrm{s}}^2\end{array}$

Hence,

The maximum value of speed and acceleration are 3.9 cm/s and 1.54 $cm/s^2$respectively.