Simple harmonic motion of a spring and mass

[stunner5000pt]

Homework Statement

A mass hanging from a spring is displaced and released so that it vibrates vertically. Its
maximum height above a tabletop is 30 cm and its minimum height above the table top
is 12 cm. The mass vibrates 20 times per minute. At time 0, its height is 30 cm.

Calculate the maximum velocity of the mass, and the times at which it occurs

Homework Equations

s(t) represents displacement
v(t) represents s'(t) which is velocity

The Attempt at a Solution

Since we start at a maximum, we must use a cosine function
the period T is 1/20 min. k = 2\pi / T = 40\pi
amplitude = (max - min) / 2 = \frac{30-12}{2} = 9 cm
s(t) = 9\cos \left(40\pi t\right) + 21
v(t) = -360\pi \sin\left(40\pi t\right)
a(t) = -1440 \pi^2 \cos\left(40\pi t\right)

we find the max velocity when we solve a(t) = 0
so we need to solve
cos \left( 40\pi t \right) = 0

since cos is negative for all half pi values, and consider only one rotation, for now...

40 \pi t = \frac{\pi}{2} \text{or} \frac{3\pi}{2}

which yields t = \frac{1}{80} \text{or} \frac{3}{80}
both answers have units of minutes.

Since the function v(t) first decreases and then increases, the critical point 3/80 is where the max occurs
if we plug each into v(t) we get
v \left(\frac{3}{80}\right) = 1131 cm/min<br /> <br /> Is this all correct? Thank you for your help!

 

[SammyS]

Homework Statement

A mass hanging from a spring is displaced and released so that it vibrates vertically. Its
maximum height above a tabletop is 30 cm and its minimum height above the table top
is 12 cm. The mass vibrates 20 times per minute. At time 0, its height is 30 cm.

Calculate the maximum velocity of the mass, and the times at which it occurs

Homework Equations

s(t) represents displacement
v(t) represents s'(t) which is velocity

The Attempt at a Solution

Since we start at a maximum, we must use a cosine function
the period T is 1/20 min. k = 2\pi / T = 40\pi
amplitude = (max - min) / 2 = \frac{30-12}{2} = 9 cm
s(t) = 9\cos \left(40\pi t\right) + 21
v(t) = -360\pi \sin\left(40\pi t\right)
a(t) = -1440 \pi^2 \cos\left(40\pi t\right)

we find the max velocity when we solve a(t) = 0
so we need to solve
cos \left( 40\pi t \right) = 0

since cos is negative for all half pi values, and consider only one rotation, for now...

40 \pi t = \frac{\pi}{2} \text{or} \frac{3\pi}{2}

which yields t = \frac{1}{80} \text{or} \frac{3}{80}
both answers have units of minutes.

Since the function v(t) first decreases and then increases, the critical point 3/80 is where the max occurs
if we plug each into v(t) we get
v \left(\frac{3}{80}\right) = 1131 cm/min<br /> <br /> Is this all correct? Thank you for your help!

<br /> It all looks correct to me.<br /> <br /> I would have been inclined to express the time in seconds rather than minutes. T = 3 sec.<br /> <br /> As far as max velocity is concerned:<br /> <br /> You know that the output of the sine function goes from -1 to 1, inclusive so if $\displaystyle \ v(t) = -360\pi \sin\left(40\pi t\right)\,,\$ then v_max = 360<span style="font-family: 'Comic Sans MS'">π . This gives an additional check regarding the fact the you have the correct time for max velocity.</span>