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Question Number 28339 by Tinkutara last updated on 24/Jan/18

	Answered by mrW2 last updated on 24/Jan/18

	$w h e n t h e r i n g i s r e l e a s e d, i t f a l l s d o w n$ $w i t h t h e m a x . a c c e l e r a t i o n (g) . d u e t o$ $t h e i n c r e a s i n g v e r t i c a l c o m p o n e n t o f$ $t h e f o r c e i n t h e s t r i n g, t h e a c c e l e r a t i o n$ $o f t h e r i n g d e c r e a s e s .$ $l e t θ = a n g l e b e t w e e n s t r i n g a n d h o r i z o n t$ $l e t a = a c c e l e r a t i o n o f r i n g$ $m_{1} g - m_{2} g \sin θ = m_{1} a$ $\Rightarrow a = g (1 - \frac{m_{2}}{m_{1}} \sin θ)$ $f o r a = 0,$ $1 = \frac{m_{2}}{m_{1}} \sin θ$ $\Rightarrow \sin θ = \frac{m_{1}}{m_{2}}$ $t h a t m e a n s : i f m_{1} < m_{2}, t h e r e i s a$ $p o s i t i o n θ_{1} = \sin^{- 1} \frac{m_{1}}{m_{2}} w h e r e t h e {r i n g}^{'} s$ $a c c e l e r a t i o n i s z e r o a n d i t s s p e e d$ $r e a c h e s t h e m a x i m u m . b u t i f m_{1} ⩾ m_{2},$ $t h e {r i n g}^{'} s a c c e l e r a t i o n r e m a i n s$ $a l w a y s > 0 a n d i t f a l l s d o w n$ $c o n t i n u o u s l y .$ $T h e r e f o r e w e h a v e f o l l o w i n g c a s e s :$ $B y (A) a n d (B), s i n c e m_{1} ⩾ m_{2}, t h e r i n g$ $f a l l s c o n t i n u o u s l y a s l o n g a s t h e$ $l e n g t h o f s t r i n g a l l o w s .$ $B y (C) w i t h m_{1} = \frac{m_{2}}{2}, w e g e t θ_{1} = 30 ° .$ $T h e m o t i o n o f t h e r i n g i s l i k e t h i s :$ $R e l e a s e d a t θ = 0, t h e r i n g f a l l s w i t h$ $a_{m a x} = g, a t θ = 30 ° i t s a = 0 a n d r e a c h e s$ $v_{m a x}, t h e n i t o b t a i n s a n e g a t i v e$ $a c c e l e r a t i o n a n d i t s s p e e d g e t s s m a l l e r$ $t i l l i t r e a c h e s t h e l o w e s t p o i n t . t h e n$ $t h e r i n g b e g i n s t o m o v e u p w a r d s b a c k$ $t o t h e o r i g i n a l p o s i t i o n θ = 0 . t h i s$ $c y c l e w i l l r e p e a t a n d r e p e a t .$ $i t m e a n s i f m_{1} < m_{2} t h e r i n g w i l l$ $m a k e a n o s c i l l a t i o n m o t i o n .$
	Commented by Tinkutara last updated on 25/Jan/18
	Is it 30° or 60° for horizontal?
	Commented by ajfour last updated on 24/Jan/18

	$t h a n k y o u S i r .$
	Commented by Tinkutara last updated on 25/Jan/18
	Thank you very much Sir! I got the answer.
	Commented by mrW2 last updated on 25/Jan/18

	$30 °$ $t h e a n s w e r i n y o u r b o o k i s 60 °, b u t {t h a t}^{'} s$ $w r o n g .$
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