A mass on a spring vibrates horizontally on a smooth level surface (see the figure). Its equation of motion is
x(t) = 8 sin t, where t is in seconds and x is in centimeters.
The first part of the question is to find the velocity and acceleration equations, which I got to be: v(t)=8cost and a(t)=-8sint.
Next, it asks: Find the position, velocity, and acceleration of the mass at time t = 2ð/3.
For v(t), I got -4, which is correct. For the x(t), I got 6.93, which is wrong, and a(t), I got -6.93, which is also wrong. What am I doing?
you v(t) and a(t) are both correct
sin 2π/3 = √3/2
cos 2π/3 = 1/2
so 8(√3/2) = 4√3 = appr 6.92820323 on my calculator
so it depends what accuracy is required
both of your answers are correct to two decimal places.
Well, it seems like you have mixed up your calculations somewhere. Let's go through it step by step:
Given the equation of motion x(t) = 8 sin(t), where t is in seconds and x is in centimeters, we need to find the position (x), velocity (v), and acceleration (a) at time t = 2π/3.
Let's start with the position:
x(t) = 8 sin(t)
x(2π/3) = 8 sin(2π/3)
x(2π/3) = 8 sin(120°)
Now, using the unit circle, we know that sin(120°) = √(3)/2.
x(2π/3) = 8 * √(3)/2
x(2π/3) = 4√(3)
So, the correct position of the mass at t = 2π/3 is 4√(3) cm.
Now, let's find the velocity:
v(t) = dx(t)/dt = d(8 sin(t))/dt = 8 cos(t)
v(2π/3) = 8 cos(2π/3)
v(2π/3) = 8 * (-1/2)
v(2π/3) = -4
So, the velocity of the mass at t = 2π/3 is -4 cm/s, which you got right!
Finally, let's find the acceleration:
a(t) = dv(t)/dt = d(8 cos(t))/dt = -8 sin(t)
a(2π/3) = -8 sin(2π/3)
a(2π/3) = -8 * √(3)/2
a(2π/3) = -4√(3)
So, the correct acceleration of the mass at t = 2π/3 is -4√(3) cm/s^2.
Looks like you made a sign error when calculating position and acceleration. Remember to double-check your calculations and pay attention to signs. Keep up the good work and don't be too hard on yourself. Math can be tricky sometimes, but you'll get the hang of it!
To find the position, velocity, and acceleration of the mass at a specific time, t = 2π/3, you need to substitute this value into the equations for x(t), v(t), and a(t).
Given:
x(t) = 8 sin(t)
v(t) = 8 cos(t)
a(t) = -8 sin(t)
Let's calculate the values at t = 2π/3:
Position, x(2π/3):
x(2π/3) = 8 sin(2π/3)
x(2π/3) = 8 sin(120°)
x(2π/3) = 8 √3/2
x(2π/3) = 4√3
So, the correct value for x(2π/3) is 4√3 or approximately 6.93 cm.
Velocity, v(2π/3):
v(2π/3) = 8 cos(2π/3)
v(2π/3) = 8 cos(120°)
v(2π/3) = 8 (-1/2)
v(2π/3) = -4
The correct value for v(2π/3) is -4 cm/s.
Acceleration, a(2π/3):
a(2π/3) = -8 sin(2π/3)
a(2π/3) = -8 sin(120°)
a(2π/3) = -8 √3/2
a(2π/3) = -4√3
So, the correct value for a(2π/3) is -4√3 or approximately -6.93 cm/s^2.
Please make sure to use the correct radian measure (2π/3) when substituting it into the trigonometric functions.
To find the position, velocity, and acceleration of the mass at a specific time, t = 2π/3, we can substitute this value into the given equations.
Given:
x(t) = 8 sin t
v(t) = 8 cos t
a(t) = -8 sin t
Let's start by finding the position x(2π/3):
x(2π/3) = 8 sin (2π/3)
To evaluate this expression, recall that sin (2π/3) = √3/2.
x(2π/3) = 8 * (√3/2) = 4√3 cm
So, the correct position of the mass at t = 2π/3 is 4√3 cm.
Next, let's find the velocity v(2π/3):
v(2π/3) = 8 cos (2π/3)
To evaluate this expression, recall that cos (2π/3) = -1/2.
v(2π/3) = 8 * (-1/2) = -4 cm/s
So, the correct velocity of the mass at t = 2π/3 is -4 cm/s.
Finally, let's find the acceleration a(2π/3):
a(2π/3) = -8 sin (2π/3)
To evaluate this expression, recall that sin (2π/3) = √3/2.
a(2π/3) = -8 * (√3/2) = -4√3 cm/s^2
So, the correct acceleration of the mass at t = 2π/3 is -4√3 cm/s^2.
It seems you made a mistake when evaluating sin (2π/3). Make sure you are using the correct value (√3/2) when calculating the position and acceleration at t = 2π/3.