Question

A person normally weighing $60kg$ stands on a platform which oscillates up and down simple harmonically with a frequency $2Hz$ and an amplitude $5cm.$ If a machine on the platform gives the person's weight, then consider the following statements :-
(i) The maximum reading of machine will be $108kg$
(ii) The maximum reading of machine will be $90kg$
(iii) The minimum reading of machine will be $12kg$
(iv) The minimum reading of the machine will be zero correct statements are :-

• A. $\left(i\right),\left(iii\right)$
• B. $\left(ii\right),\left(iii\right)$
• C. $\text{(i), (iv)}$
• D. $\text{(ii), (iv)}$

Answer 1

The correct option is A $\left(i\right),\left(iii\right)$

R.E.F image

Solution

we know in SHM,

$a_{max}=A{w}^2$ and $a_{min}=-A{w}^2$

Here,

$a_{max}=\left(\frac{5}{100}\right)(2\times 2\pi {)}^2,a_{min}=(\frac{-5}{100}\left)\right(2\times 2n{)}^2$

$a_{max}=8m/s^2$ and $a_{min}=-8m/s^2$

FBD will be

$N-Mg=Ma$

$N=M(g+a)$

As, wight $=N=M(g+a)$

Max.wight $=M(g+a_{max})$

$W_{max}=M(9.8+8)$

$W_{max}=M\left(17.8\right)=60\times 178=106.8N$

Min weight $=M(g+a_{min})$

$W_{min}=M(9.8-8)=60\times 1.8=108N$

Measured Max Mass $=\frac{W_{max}}{g}=\frac{1068}{9.8}=108kg$

Measured Min Mass $=\frac{W_{min}}{g}=\frac{108}{9.8}=12kg$

Option - A is correct