Given the equation for simple harmonic method d=1/12 sin(60 pi t) answer the following questions:
b)Find the frequency
i got 60pi
c)Calculate the value of d when t=3
d)Determine the least positive value of t for which d=0.
b) correct, the period would be 2π/60π = 1/30
c) when t = 3
d = (1/12)sin (180π) = 0
d) sin(60πt) = 0 when
t = 0 , 1/30, 1/15 ...
so the first positive value is 1/30
b) To find the frequency, we need to identify the constant coefficient in front of the "t" term in the equation. In this case, the coefficient is 60π. The frequency of the simple harmonic motion is given by the formula:
Frequency = (1 / (2π)) * coefficient
In this case, the frequency would be:
Frequency = (1 / (2π)) * 60π = 30 Hz
c) To calculate the value of d when t = 3, we substitute the value of t into the equation:
d = (1/12) * sin(60π * 3)
= (1/12) * sin(180π)
= (1/12) * 0
= 0
Therefore, when t = 3, the value of d is 0.
d) To determine the least positive value of t for which d = 0, we need to find the time at which the sine function in the equation becomes 0. Since sin(0) equals 0, we can set the argument of the sine function equal to 0 and solve for t:
60πt = 0
Dividing both sides by 60π, we get:
t = 0
Therefore, the least positive value of t for which d = 0 is t = 0.
To find the frequency of the equation, we need to identify the coefficient of \(t\) in the equation. In this case, the coefficient is \(60\pi\).
b) The frequency (\(f\)) can be found using the formula:
\[f = \frac{\text{coefficient of } t}{2\pi}\]
Plugging in the coefficient, we get:
\[f = \frac{60\pi}{2\pi} = 30\text{ Hz}
c) To calculate the value of \(d\) when \(t = 3\), substitute \(t = 3\) into the equation:
\[d = \frac{1}{12}\sin(60\pi(3))\]
Simplifying this expression, we get:
\[d = \frac{1}{12}\sin(180\pi)\]
Since \(\sin(180\pi) = 0\), we can further simplify:
\[d = \frac{1}{12}(0) = 0
d) To determine the least positive value of \(t\) for which \(d = 0\), we need to solve the equation:
\[0 = \frac{1}{12}\sin(60\pi t)\]
Since \(\sin(60\pi t)\) will be equal to zero when the angle \(\sin^{-1}(0)\) is a multiple of \(\pi\), we set \(60\pi t\) equal to \(k\pi\), where \(k\) is an integer:
\[60\pi t = k\pi\]
Solving for \(t\), we get:
\[t = \frac{k}{60}\]
The least positive value of \(t\) when \(d = 0\) is \(t = \frac{1}{60}\) (taking \(k = 1\)).
Therefore, the answers are:
b) The frequency is 30 Hz.
c) The value of \(d\) when \(t = 3\) is 0.
d) The least positive value of \(t\) for which \(d = 0\) is \(t = \frac{1}{60}\).