A mass m, attached to a spring with constant k, oscillates...

Name: A mass m, attached to a spring with constant k, oscillates according to x(t) = A sin(wt) where w^2 = k/m. When x = A/2, what is the ratio of the kinetic energy KE/Etot of the spring-mass system? 3/4 1/4 1 1/2
Uploaded: 2022-07-13T14:39:56-08:00
Duration: 5 min 58 s
Channel: Adi S
Description: A mass m, attached to a spring with constant k, oscillates according to x(t) = A sin(wt) where w^2 = k/m. When x = A/2, what is the ratio of the kinetic energy KE/Etot of the spring-mass system? 3/4 1/4 1 1/2

A mass m, attached to a spring with constant k, oscillates according to x(t) = A sin(wt) where w^2 = k/m. When x = A/2, what is the ratio of the kinetic energy KE/Etot of the spring-mass system? 3/4 1/4 1 1/2

Transcript

00:01 Here in the question they have given the position vector is x of t is equal to a sine omega t let it be the first equation by doing the derivative of this position vector we can get the velocity vector because velocity is nothing but the rate of change of position so d by d t of a sine omega here a is a constant so we need there is no derivative for other constants and d by d t of sine omega 3 we got cos omega t and again the derivative for the omega t becomes omega so a omega cose omega t is the velocity vector so let it be equation two and in the question itself they have given one condition that is when x is equal to a by 2 x of t is equal to a sine omega t so that implies a by 2 is equal to a sine omega t both sides a get cancelled half is equal to sine omega t here half we can write in the form of sign functions that is sine pi by six value is 1 by 2 is equal to sine omega so the angle omega t is equal to pi by 6 let it be equation 3 now we can calculate the kinetic energy and potential energy for calculating total energy the kinetic energy formula we know that is half m v square v of t whole square and half m velocity vector is a omega cos omega t whole square and there is a condition that is at x is equal to a by 2 omega t value as pi by 6 by substituting this kinetic energy is equal to half m a square omega square cos omega t minus pi by 6 whole square that is nothing but cos square y by 6 half m a square omega square cos square pi by six value cost pi by six value is root 3 by 2 by by applying the square we'll get 3 by 4 so 3 by 8 m a square omega square is the kinetic energy then after we can calculate potential energy also potential energy is equal to half kx square here we can substitute k value as m omega square because omega square is equal to k by m this is the condition they have given in the question so from this k is equal to m omega square half m omega square and x is nothing but a position vector a sine omega t whole square so half m omega square a square sine omega t value as pi by six half m omega square a square sign pi by six sine pi by six value as one by two so one by two whole square is one by so the potential energy is one by eight m omega square a square so therefore the total energy is equal to potential energy plus kinetic energy.

04:25 Potential energy is 1 by 8 m omega square a square plus kinetic energy is 3 by 8 m omega square a square.

04:40 So that is equal to taking the 8 as an lcm 3 plus 1 becomes 4m omega square a square by 8.

04:47 That will be half m omega square a square...

Question

A mass m, attached to a spring with constant k, oscillates according to x(t) = A sin(wt) where w^2 = k/m. When x = A/2, what is the ratio of the kinetic energy KE/Etot of the spring-mass system? 3/4 1/4 1 1/2

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