Question

please write answer with process​

Answer 1

Answer:

Height difference of the spring is

$(l - lcos \theta) \\ l \: is \: the \:initial \: length \\ and \: lcos \theta \: is \: final \: length \\ so \: the \: difference \: will \: be \: its \: height \\ \: difference$

$(upward)f = kx \\ kxcos \theta = k(l - lcos \theta)$

then subject the x value...the you will get

$x = \frac{l}{cos \theta} - l \\ which \: is \: also \: equal \: to \\ x = lsec \theta - 1$

this is our 1st equation

$x = lsec \theta - l$

then we use this below equation as our 2nd equation

$kxcos( \theta) = mg$

then substitute that (x) we got from the previous equation to the above equation

$kxcos \theta = mg \\ k(lsec \theta - l) cos \theta= mg \\ (klsec \theta - kl)cos \theta= mg \\ klsec \theta.cos \theta - klcos \theta = mg \\ \frac{kl}{cos \theta} \times cos \theta - klcos \theta = mg$

then subject the cos(theta)...then you get

$cos \theta = 1 - \frac{mg}{kl} \\ cos \theta = \frac{kl - mg}{kl} \\ sec \theta = \frac{kl}{k - mg}$

special point

you know that

$1 + {tan}^{2} \theta = {sec}^{2} \theta$

so subject the Tan vale here...then you get

$tan \theta = \sqrt{ {sec}^{2} \theta - 1 }$

then substitute the sec(theta) value we get in previous equation.....

$tan \theta = \sqrt{ ({ \frac{kl}{kl - mg}) }^{2} } - 1 \\ = \sqrt{ \frac{ {k}^{2} {l}^{2} - {(kl - mg)}^{2} }{ {(kl - mg)}^{2} } }$

then take separate square root for denominator and numerator

then you know squaroot cuts with the square terms

therefore you get

$tan \theta = \frac{ \sqrt{ {k}^{2} {l}^{2} - {(kl - mg)}^{2} } }{kl - mg}$

look at the 1st pic attached...

therefore

$ltan( \theta) = s$

then substitute the tan( theta) value we

got in previous eqaution to the above equation

therefore

$ltan \theta = s \\ s = l \frac{ \sqrt{ {k}^{2} {l}^{2} - {(kl - mg)}^{2} } }{kl - mg}$

so hope you can understand thank you