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Asked by Prashant DIGHE | 22 May, 2020, 12:44: PM

Expert Answer

From the given data,

one end is fixed and other is open.

So at node, when x =0 y(t) = 0

At antinode, when x = 3 m and

y left parenthesis t right parenthesis space equals space plus-or-minus space A

velocity, v = 100 m/s

and we know,

v = ω/k

S o space f o r space b e i n g space t h e space p o s s i b l e space w a v e f o r m comma space t h e space g i v e n space o p t i o n s space m u s t space s a t i f y space t h e space a b o v e space m e n t i o n e d space c o n d i t i o n s. space L e t apostrophe s space c h e c k space f o r space o p t i o n space left parenthesis A right parenthesis y space left parenthesis t right parenthesis space equals space A space sin space pi over 6 space cos fraction numerator space 50 pi t over denominator 3 end fraction space N o w space s u b s i t u t i n g space x space equals 0 space i n space t h e space e q u a t i o n space w e space g e t space y left parenthesis t right parenthesis space equals 0 space O n space s u b s t i t u t i n g space x equals 3 space space w e space g e t comma space y space equals space fraction numerator A cos 50 pi t over denominator 3 end fraction space equals space A space left parenthesis f o r space t space equals 0 right parenthesis space omega space equals space fraction numerator 50 space pi over denominator 3 end fraction comma space k space equals pi over 6 S u b s t i t u t i n g space i n comma space v space equals omega over k space equals space fraction numerator begin display style bevelled fraction numerator 50 pi over denominator 3 end fraction end style over denominator begin display style bevelled pi over 6 end style end fraction space equals space 100 space m divided by s S o space o p t i o n space left parenthesis A right parenthesis space s a t i s f i e s space a l l space t h e space c o n d i t i o n s. space S o space i t space c a n space b e space t h e space p o s s i b l e space w a v e f o r m space o f space t h e space s t a t i o n a r y space w a v e. space I n space s i m i l a r space w a y space l e t apostrophe s space c h e c k space f o r space r e s t space o f space t h e space o p t i o n s. O p t i o n space left parenthesis B right parenthesis thin space space i s space y left parenthesis t right parenthesis space equals space A space sin space fraction numerator pi x over denominator 3 end fraction space cos fraction numerator 100 pi t over denominator 3 end fraction space w h e n space x space equals 0 comma space y left parenthesis t right parenthesis equals 0 w h e n space x space equals 3 space t h e n space y space equals 0 space space t h u s space t h e space g i v e n space c o n d i t i o n space i s space o t space s a t i s f i e d space b y space t h i s space o p t i o n. space O p t i o n space left parenthesis C right parenthesis space i s space y left parenthesis t right parenthesis space equals space space A space sin space fraction numerator 5 pi x over denominator 6 end fraction space cos fraction numerator space 250 pi t over denominator 3 end fraction space w h e n space x space equals 0 comma space y space equals 0 space w h e n space x space equals 3 comma space y space equals space A space v space equals space omega over k space equals space fraction numerator begin display style bevelled fraction numerator 250 pi over denominator 3 end fraction end style over denominator begin display style bevelled fraction numerator 5 pi over denominator 6 end fraction end style end fraction space equals space 100 space m divided by s space T h u s comma space a l l space t h e space c o n d i t i o n s space a r e space s a t i s f i e d space b y space t h i s space o p t i o n space left parenthesis C right parenthesis. space T h u s comma space i t space c a n space b e space t h e space p o s s i b l e space w a v e f o r m N o w comma space O p t i o n space left parenthesis D right parenthesis space i s space y left parenthesis t right parenthesis space equals space space A space sin space fraction numerator 5 pi x over denominator 2 end fraction space cos space 250 pi t w h e n space x space equals 0 comma space y space equals 0 space x space equals 3 space a n d space y space equals negative A space v space equals fraction numerator begin display style 250 pi end style over denominator begin display style bevelled fraction numerator 5 pi over denominator 2 end fraction end style end fraction space equals space 100 space m divided by s space T h u s comma space o p t i o n space left parenthesis D right parenthesis space a l s o space s a t i s f i e s space t h e space c o n d i t i o n s space a s space p e r space t h e space g i v e n space i n f o r m a t i o n.

Hence, option (A), (C) and (D) are correct.

Answered by Shiwani Sawant | 22 May, 2020, 10:00: PM

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