# A 0.78 kg block is attached to a horizontal spring of force constant 256 kg/s2, resting atop a frictionless surface. The block is pulled back 30 cm and then released. What is the period of the oscillations of the block? Maximum speed of the oscillation?

$0.347 s$ & $5.435 \setminus \frac{m}{s}$

#### Explanation:

The period of oscillation of spring-mass system on any frictionless plane (horizontal, vertical or inclined) is given as follows

$T = 2 \setminus \pi \setminus \sqrt{\frac{m}{k}}$

Where, $m$ mass of system/block & $k$ spring constant of spring

Hence, setting $m = 0.78$ kg & $k = 256 \setminus {\textrm{k \frac{g}{s}}}^{2}$ in above formula, the period of oscillation is given as follows

$T = 2 \setminus \pi \setminus \sqrt{\frac{0.78}{256}} = 0.347 \setminus \setminus \textrm{\sec}$

Now, the maximum speed ${v}_{\setminus \textrm{\max}}$ of block of mass $m = 0.78$ kg will be at mean position when entire elastic energy of spring $= \frac{1}{2} k {x}^{2}$ is transferred to the block when spring is given a displacement $x = 30$ cm

$\frac{1}{2} m {v}_{\setminus \textrm{\max}}^{2} = \frac{1}{2} k {x}^{2}$

$\frac{1}{2} \left(0.78\right) {v}_{\setminus \textrm{\max}}^{2} = \frac{1}{2} \left(256\right) {\left(0.3\right)}^{2}$

${v}_{\setminus \textrm{\max}} = 5.435 \setminus \frac{m}{s}$