Respuesta :
Answer:
Position of the object at t = 5 s is [tex]9.42[/tex] centimeter
Explanation:
The position of object as a function of time can be represented as
[tex]X _t = A Sin (\omega t + \theta)[/tex]
It is given that at t [tex]=0[/tex] seconds the position of the object is [tex]- 5[/tex] centimeter.
We will substitute these values to determine the value of [tex]\theta[/tex]
[tex]- 5 = 10 sin (\omega t + \theta)\\- 5 = 10 sin (\omega * 0 + \theta)\\- 5 = 10 sin ( (\theta)\\sin ( \theta) = \frac{-5}{10} =\frac{-1}{2} \\ ( \theta) = sin^{-1} (\frac{-1}{2})\\ ( (\theta) = -0.52 \\(\theta) = 3.66[/tex]
Velocity at initial stage when t [tex]=0[/tex] seconds is [tex]3 \frac{cm}{s}[/tex]
[tex]3 \frac{cm}{s}[/tex] [tex]= 10 * \omega * cos (-0.52)\\[/tex]
[tex]\omega = 3.5 \frac{rad}{s}[/tex]
Position after [tex]5[/tex] seconds
[tex]X = 10 * Sin(\omega t + \theta)[/tex]
[tex]X= 10 * sin ( 3.5 * 5 -0.52 )\\X = 9.42[/tex]
The position (in cm) of the object which oscillates in 1-dimension, at t = 5 s is 9.42 cm.
What is position function?
The position function helps to find the position of the object after the time t in sine and cosine function of simple harmonic motion.
The position sine function of an object can be given as,
[tex]X(t)=A\sin(\omega t+\theta)[/tex]
An object oscillates in 1-dimension with an amplitude of 10 cm. At t = 0 s, its position is x = -5 cm and it is moving with a velocity of 3 cm/s.
Put the values in the above formula as,
[tex]-5=10\sin(\omega\times0+\theta)\\\theta=\sin^{-1}(\dfrac{-5}{10})=0.52\rm rad\\[/tex]
When the value of time is 0, then the velocity is 3 cm/s. Therefore, by the cosine equation,
[tex]3=10\times \omega \cos(-0.52)\\\omega=3.5\rm rad/s[/tex]
Hence, the position after the 5can be given as,
[tex]X=10\sin(\omega t+\theta)\\X=10\sin(3.5\times5+(-0.52))\\X=9.42\rm cm[/tex]
Hence, the position (in cm) of the object which oscillates in 1-dimension, at t = 5 s is 9.42 cm.
Learn more about the position function here;
https://brainly.com/question/24786395
