- #1
DuoASM
Hi guys, I would be grateful if you could help me out with this problem, I've worked though it myself, but my answer to part c is very large, which is making me doubt my answer, and in turn making me doubt my answer to part b
The midpoint of a guitar string executes s.h.m with a motion following the form:
$$x(t) = A\ \textrm{sin}(\omega t + \Phi)$$
where:
b)What is the maximum speed of the string during this motion?
c)what is the maximum magnitude of the acceleration of the string?
d) Determine the initial displacement, velocity and acceleration of the midpoint of the spring
$$x(t) = A\ \textrm{sin}(\omega t + \Phi)$$
$$v(t) = A\omega\ \textrm{cos}(\omega t + \Phi)$$
$$a(t) = -A\omega^2\ \textrm{sin}(\omega t + \Phi)$$
$$ T = \frac{2\pi}{\omega}$$
a) using $$ T = \frac{2\pi}{\omega}$$
$$ T = \frac{2\pi}{2.76 \times 10^3 s^{-1}} = 2.28 \times 10^{-3}s$$
b) Maximum speed is obtained when $$\textrm{cos}(\omega t + \Phi) = 1$$
So maximum speed is given by: $$ A\omega \times 1 = 1.60 \times 10^{-3}m \times 2.76 \times 10^3 s^{-1} =4.42ms^{-1} $$
c) Similarly, maximum magnitude of acceleration is found when $$\textrm{sin}(\omega t + \Phi) = 1$$
So: $$ |a| =A\omega^2 = 1.60 \times 10^{-3}m \times (2.76 \times 10^3 s^{-1})^2 = 1.22 \times 10^4 ms^{-2}$$
d) The initial values are found at $$ t = 0 $$ so it follows that $$ \omega t = 0$$
so, for the initial displacement: $$x(t) = A\ \textrm{sin}(\omega t + \Phi)$$
$$ x(0) =1.60 \times 10^{-3} m \ \textrm{sin}(0 + \frac{\pi}{2}) = 1.60 \times 10^{-3} m$$
For initial velocity, as $$ \textrm{cos}(0 + \frac{\pi}{2}) = 0$$
then it follows that: $$v(0) = A\omega\ \textrm{cos}( 0+ \frac{\pi}{2}) = 0$$
And for the initial acceleration:
$$a(t) = -A\omega^2\ \textrm{sin}(\omega t + \Phi)$$
$$a (0)= 1.60 \times 10^{-3}m \times (2.76 \times 10^3 s^{-1})^2 \textrm{sin}(0 + \frac{\pi}{2})$$ $$a(0)= 1.22 \times 10^4 ms^{-2}$$I hope the formatting is okay, its the first time I have tried using LaTex. If there is anything wrong please point it out, it will all help me improve. Thank you for your time in advance!
Homework Statement
The midpoint of a guitar string executes s.h.m with a motion following the form:
$$x(t) = A\ \textrm{sin}(\omega t + \Phi)$$
where:
- $$A = 1.60mm =1.60 \times 10^{-3} m$$
- $$\omega = 2.76 \times 10^3 s^{-1}$$
- $$\Phi = \frac{\pi}{2}$$
b)What is the maximum speed of the string during this motion?
c)what is the maximum magnitude of the acceleration of the string?
d) Determine the initial displacement, velocity and acceleration of the midpoint of the spring
Homework Equations
$$x(t) = A\ \textrm{sin}(\omega t + \Phi)$$
$$v(t) = A\omega\ \textrm{cos}(\omega t + \Phi)$$
$$a(t) = -A\omega^2\ \textrm{sin}(\omega t + \Phi)$$
$$ T = \frac{2\pi}{\omega}$$
The Attempt at a Solution
a) using $$ T = \frac{2\pi}{\omega}$$
$$ T = \frac{2\pi}{2.76 \times 10^3 s^{-1}} = 2.28 \times 10^{-3}s$$
b) Maximum speed is obtained when $$\textrm{cos}(\omega t + \Phi) = 1$$
So maximum speed is given by: $$ A\omega \times 1 = 1.60 \times 10^{-3}m \times 2.76 \times 10^3 s^{-1} =4.42ms^{-1} $$
c) Similarly, maximum magnitude of acceleration is found when $$\textrm{sin}(\omega t + \Phi) = 1$$
So: $$ |a| =A\omega^2 = 1.60 \times 10^{-3}m \times (2.76 \times 10^3 s^{-1})^2 = 1.22 \times 10^4 ms^{-2}$$
d) The initial values are found at $$ t = 0 $$ so it follows that $$ \omega t = 0$$
so, for the initial displacement: $$x(t) = A\ \textrm{sin}(\omega t + \Phi)$$
$$ x(0) =1.60 \times 10^{-3} m \ \textrm{sin}(0 + \frac{\pi}{2}) = 1.60 \times 10^{-3} m$$
For initial velocity, as $$ \textrm{cos}(0 + \frac{\pi}{2}) = 0$$
then it follows that: $$v(0) = A\omega\ \textrm{cos}( 0+ \frac{\pi}{2}) = 0$$
And for the initial acceleration:
$$a(t) = -A\omega^2\ \textrm{sin}(\omega t + \Phi)$$
$$a (0)= 1.60 \times 10^{-3}m \times (2.76 \times 10^3 s^{-1})^2 \textrm{sin}(0 + \frac{\pi}{2})$$ $$a(0)= 1.22 \times 10^4 ms^{-2}$$I hope the formatting is okay, its the first time I have tried using LaTex. If there is anything wrong please point it out, it will all help me improve. Thank you for your time in advance!
