A question about Simple Harmoic Motion

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To find the period T of the simple harmonic motion described by Z(t)=(5)Sin(2∏/3t), the constant multiplier in the sine function is identified as 2∏/3. The standard period of a sine function is 2π, which is modified by the constant K, resulting in a period calculated as T = 2π/(2π/3). This simplifies to T = 3 seconds. Therefore, the period of the motion is 3 seconds.
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Homework Statement



Let Z(t)=(5)Sin(2∏/3t) in SHO

guys, who knows how to find the Period T from above equation ?

thx
 
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Read this: http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html"
 
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ooooooo... let's see...

The constant in this sin function is 2*pi/3 right? Since a regular sin function has a period of 2 pi and a constant K multiplier reduces the period to 2pi/K, and K in your situation is 2*pi/3... then...
 
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