Question

block of mass 1µg is connected with an elastic string of stiffness constant K = 10–5 N/m. Now a light pulse of intensity I = 20 W/m2 strikes the block at its vertical surface of surface area 10 cm2 at an angle 53° with horizontal. If surface of block is 100% absorbing the light and the duration of light pulse is 6 ms then the max displacement of block from it’s mean position is N µm. Find value of N.

Answer 1

Given that,

Mass = 1μg

Stiffness constant $K=10^{-5}\ N/m$

We know that,

Maximum displacement :

The maximum displacement is called amplitude.

We need to calculate the maximum displacement

Using restoring force

$F=-kx$

Negative sign shows the opposite direction of motion

$x=\dfrac{mg}{k}$

Where, k = spring constant

m = mass

g = acceleration due to gravity

Put the value into the formula

$x=\dfrac{1\times10^{-9}\times9.8}{10^{-5}}$

$x =0.00098\ m$

$x=980\ \mu m$

The max displacement of block from its mean position is N μm.

So, The value of N is 980.

Hence, The maximum displacement is 980μm.