Dear Student let y=2sin(0 1πx)cos100πt cmthis can be written as y=2cos(100πt)sin(0 1πx)amplitudeA=2cos(100πt)equation is y=Asinkxso k=2πλ=0 1πλ=20 1=20 cmdistance between the node and next antinode is=λ/4=20/4=5 cmoption B is right Regards

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Solve is located at 3m. the next node uill be located at x is equal to a) 3.25m b) 3.75m "3.50m d)4m Ifthe particle displacement in a standing wave is given byy=2 sin (0.1 It x)cos 100% tcm, the distance between a node and the next antinode v.ill be a) 2.5 cm b)5cm c) 7.5cm d) lhe simplest mode ofa vibration ofthe string is —-

Dear Student

l e t y = 2 \sin (0.1 π x) \cos 100 π t c m t h i s c a n b e w r i t t e n a s y = 2 \cos (100 π t) \sin (0.1 π x) a m p l i t u d e A = 2 \cos (100 π t) e q u a t i o n i s y = A \sin k x s o k = \frac{2 π}{λ} = 0.1 π λ = \frac{2}{0.1} = 20 c m d i s \tan c e b e t w e e n t h e n o d e a n d n e x t a n t i n o d e i s = λ / 4 = 20 / 4 = 5 c m o p t i o n B i s r i g h t

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