Acceleration in Harmonic Motion

[ciao_potter]

Problem:
A pendulum that has a period of 3.00000s and that is located where the accleration due to gravity is 9.79 m/s^2 is moved to a location where the acceleration due to gravity is 9.82 m/s^2. What is its new period, in s?

Equations
Equation for Harmonic motion: x = A sin (2pi * f * t)

Acceleration for Harmonic motion: a = -4pi^2 A f^2 sin(2pi f t)

f = 1/t = 1/3

Attempt:
Let A = 4 for both problems.
9.79 = 16/9 * pi^2 * sin (2pi/3 t)

Solving for t,
sin(2pi/3 * t) = 0.55796309317
2pi/3 * t = 0.5919292673 (radians)
t = 0.282654065 seconds

9.82 = 16 * pi^2 * f^2 sin(2pi * 0.282654 * f)
f^2 sin(2pi * 0.282654 * f) = 0.06218587646

Now I get a lot of answers for f.
https://www.wolframalpha.com/input/?i=x^2+sin(2pi+*+0.282654+*+x)+=+0.06218587646

Not sure which one is the answer.

Thank you!

 

[SammyS]

Problem:
A pendulum that has a period of 3.00000s and that is located where the accleration due to gravity is 9.79 m/s^2 is moved to a location where the acceleration due to gravity is 9.82 m/s^2. What is its new period, in s?

Equations
Equation for Harmonic motion: x = A sin (2pi * f * t)

Acceleration for Harmonic motion: a = -4pi^2 A f^2 sin(2pi f t)

f = 1/t = 1/3

Attempt:
Let A = 4 for both problems.
9.79 = 16/9 * pi^2 * sin (2pi/3 t)

Solving for t,
sin(2pi/3 * t) = 0.55796309317
2pi/3 * t = 0.5919292673 (radians)
t = 0.282654065 seconds

9.82 = 16 * pi^2 * f^2 sin(2pi * 0.282654 * f)
f^2 sin(2pi * 0.282654 * f) = 0.06218587646

Now I get a lot of answers for f.
https://www.wolframalpha.com/input/?i=x^2+sin(2pi+*+0.282654+*+x)+=+0.06218587646

Not sure which one is the answer.

Thank you!

Hello ciao_potter . Welcome to PF !

You are not to use the acceleration resulting from harmonic motion.
The pendulum is not likely to have an acceleration of g .

What is the formula for the period of simple pendulum?