How Do You Calculate Frequency and Period in Simple Harmonic Motion?

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The discussion focuses on calculating frequency and period in simple harmonic motion using the provided position equation x=(4.00m) cos(3.00πt+π). The user initially calculated the period (T) as 6.58 seconds and frequency (f) as 0.15 Hz, but these values were incorrect. The error was identified in the calculation of T, where T should be derived from T=2π/ω, leading to T=2/3 seconds instead of 6.58 seconds. Additionally, the user sought help with determining the time it takes for a mass-spring system to move from a specific position, encountering difficulties with the phase constant and angular frequency. The discussion highlights the importance of correctly applying formulas in simple harmonic motion.
Paymemoney
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Hi
I have a few questions i need some help on.

Homework Statement


The position of a particle is given by the expression x=(4.00m) cos(3.00\pit+\pi), where x is in meters and t in seconds. Determine (a) the frequency and period of the motion.

Homework Equations


f=\frac{1}{T}

T=\frac{2\pi}{\omega}

The Attempt at a Solution



i got:

T=\frac{2\pi}{\omega}

T=\frac{2\pi}{3\pi}

T=6.58s

f=\frac{1}{6.5797}

f=0.15Hz

This is incorrect according to book's answer, what is wrong?

Homework Statement


A 0.500kg object attached to a spring with a force constant of 8.00N/m vibrates in simple harmonic motion with an amplitude of 10.0cm. Calculate time it takes the object to move from x=+8.0cm (just after it is released at x=+10cm) to x=0.

The Attempt at a Solution


i tried to use this formula: x=Acos(kt+\theta), however i cannot find t because there is two unknowns.

Instead what should i use?


P.S
 
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Paymemoney said:
The position of a particle is given by the expression x=(4.00m) cos(3.00\pit+\pi), where x is in meters and t in seconds.

t does not appear in the formula!

Paymemoney said:
A 0.500kg object attached to a spring with a force constant of 8.00N/m vibrates in simple harmonic motion with an amplitude of 10.0cm. Calculate time it takes the object to move from x=+8.0cm (just after it is released at x=+10cm) to x=0.

The Attempt at a Solution


i tried to use this formula: x=Acos(kt+\theta), however i cannot find t because there is two unknowns.



What is k? is it the angular frequency? If so, how is the angular frequency of the vibration related to the force constant and mass?

If the object is released at x=10 cm which is the maximum displacement, what is the value of the phase constant theta?

ehild
 
oops let me write it out again:

x=Acos(\omega t + \theta)

this is the correct equation i used

so i sub in values and i get two unknowns

8 = 10cos(4t+ \theta)
 
The time is measured from the instant when the object is released. The object is released when x=10. So what is the value of theta?

ehild
 
ok so this is what i have done:

x=Acos(\omega t + \theta)

x = 10cm, A = 10cm, \omega = 4m/s t = ? \theta = 0.93 degrees

10 = 10cos(4t + 0.93)

10 = 10cos(4t) +10cos(0.93)

10 - 10cos(0.93) = 10cos4t

t = \frac{arccos\frac{10-10cos(0.93)}{10}}{4}

t = 0.289 s
 
Could anyone tell me what is the problem with my first question??
 
Paymemoney said:
T=\frac{2\pi}{\omega}

T=\frac{2\pi}{3\pi}

T=6.58s

f=\frac{1}{6.5797}

f=0.15Hz

This is incorrect according to book's answer, what is wrong?

T=\frac{2\pi}{3\pi} reduces down to 2/3 which is not equal to 6.58.
 
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