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An object moves according to the simple Harmonic oscillator equation d2x/de​2 = -x

Show that x = Asint + Bcost is a valid solution
Show that x = Csin(t+phi) is a valid solution
Show that both of those general solutions are in fact the same by expressing A and B in terms of C and phi or vice-versa.

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Final answer:

To show that x = Asin(t) + Bcos(t) and x = Csin(t+phi) are valid solutions, we can substitute each equation into the simple harmonic oscillator equation and show that they satisfy the equation. By using trigonometric identities, we can express A and B in terms of C and phi or vice-versa.

Step-by-step explanation:

To show that x = Asin(t) + Bcos(t) is a valid solution, we need to substitute it into the simple harmonic oscillator equation d^2x/dt^2 = -x and show that it satisfies the equation. Taking the second derivative of x = Asin(t) + Bcos(t) with respect to time, we get -Asin(t) - Bcos(t), which is equal to -x. Therefore, x = Asin(t) + Bcos(t) is indeed a valid solution.

To show that x = Csin(t+phi) is a valid solution, we can apply a similar process. Taking the second derivative of x = Csin(t+phi) with respect to time, we get -Csin(t+phi), which is also equal to -x. Therefore, x = Csin(t+phi) is a valid solution.

To express A and B in terms of C and phi, we can use trigonometric identities. Based on the equation x = Asin(t) + Bcos(t), we can rewrite it as x = Rsin(t+alpha), where R = sqrt(A^2 + B^2) and alpha = atan(B/A). Comparing this with x = Csin(t+phi), we can see that C = R and phi = alpha. Therefore, C = sqrt(A^2 + B^2) and phi = atan(B/A).

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