Plus One Physics Chapter 14 | Oscillations | Sure Questions

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A 2kg body performs shm of amplitude 20cm and the restoring force acting at that position is 50N. Find the acceleration and kinetic energy of the particle when the displacement is 10cm. ee questionil answer 3.75 alle. F=kx k=F/x = 50/0.2 = 250N/m k=m(omega)^2 omega^2 = k/m = 250/2 = 125 a = -(omega)^2x = -125 * 0.1 = -12.5 m/s^2 continue the rest for K.E

[@mydreamisiit]

U are right every one please correct it

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Bro you had a mistake... K is not 250. It's 500 ( X is not 0.2, it's 0.1)

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@@hazeenajamal190 bro you had a mistake... we don't know the force or acceleration when x = 0.1m. so the k is 250.

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A 2kg body performs shm of amplitude 20cm and the restoring force acting at that position is 50N. Find the acceleration and kinetic energy of the particle when the displacement is 10cm. ee questinte answer 3.76 J alle [answer key shows 3.76J]

[@mydreamisiit]

You are right Miss paranja answer wrong aan because she used the equation F=ma ഇവിടെ ആ equation അല്ല വരുക. F = kx ആണ് എടുക്കേണ്ടത് Then acceleration = 12.5 m/s w²=k/m aan ath use cheyth w=5√5 എന്ന് കിട്ടും..and KE = 1/2k(A²-x²) equation aan edukkuka then we get ~ 3.75J Every one please correct it

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@@mydreamisiit F=Kx eduthal k engana kittum k=mw^2alle w engane kittum

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That answer is wrong oscilllations newtons equation for force edukkoola F=ma❌ F= -kx aan . Then acceleration 25m/s okke enganeya kittuka just think ...acceleration =12.5m/s aan K.E = 3.75 J (by using the equation KE=1/2k(A²-x²) Every one please correct it Arum F=ma use cheyth oscillations le equation cheyyaruth ath translational motion maathramulla equation aaan Here restoring force is considered so a is not = to k a = -(k/m)x aan Please reply eduport team

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k ethreya kittiye

[@mydreamisiit]

@@lachusz. 250N/m

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F=ma can use here and your answer is correct but what teacher made mistake is that she found the acceleration when the displacement is 20cm,but in question it is asked at displacement equal to 10cm

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13:36 , slide number 5l correction und tto F=kx k=F/x = 50/0.2 = 250N/m k=m(omega)^2 omega^2 = k/m = 250/2 = 125 a = (omega)^2x ( no need of negative as we are finding only displacement) = 25 * 0.1 = 12.5 m/s^2 KE = 1/2 mv^2 use v^2 = omega^2(A^2 -x^2) KE= 1/2 m * omega^2(A^2 -x^2) = 0.5*2*125*(0.2^2 - 0.1^2) =3.75 J

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F/x x in etha 0.2 eduthe ath amplitude ale?

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@@gayathrikrishna1048 amplitude maximum displacement ale.. 0.2 m is displaced when 50N is applied on a body with mass 2kg

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