1. A physics department has a Foucault pendulum, a long-period pendulum suspende

Question

1. A physics department has a Foucault pendulum, a long-period pendulum suspended from the ceiling. The pendulum has an electric circuit that keeps it oscillating with a constant amplitude. When the circuit is turned off, the oscillation amplitude decreases by 50% in 37 minutes.

a. What is the pendulum's time constant?

b. How much additional time elapses before the amplitude decreases to 25% of its initial value?

2. Vision is blurred if the head is vibrated at 29 Hzbecause the vibrations are resonant with the natural frequency of the eyeball held by the musculature in its socket.

If the mass of the eyeball is 7.2 g, what is the effective spring constant of the musculature attached to the eyeball?

3. As we've seen, astronauts measure their mass by measuring the period of oscillation when sitting in a chair connected to a spring. The Body Mass Measurement Device on Skylab, a 1970s space station, had a spring constant of 606 N/m. The empty chair oscillated with a period of 0.842 s .

What is the mass of an astronaut who oscillates with a period of 2.28 s when sitting in the chair?

4. The ultrasonic transducer used in a medical ultrasound imaging device is a very thin disk(m = 0.10 g) driven back and forth in SHM at1.0 MHz by an electromagnetic coil.

a. The maximum restoring force that can be applied to the disk without breaking it is 38,000 N. What is the maximum oscillation amplitude that won't rupture the disk?

b.What is the disk's maximum speed at this amplitude?

Explanation / Answer

Here is how you get the answer for the first part of the problem!

t=37 mins -> 2220sec
We want "T" which is the pendulum time constant

We want "T" which is the pendulum time constant

Using this equation
.5A=Ae^(-t/T)

The .5A is half the amplitude

Take ln of both sides to get ride of Ae
=ln(.5)=-2220/T

Now rearrange to = T
T=-2220/ln(.5) = 3202.78sec / 60 secs = 53.38 mins -> first part of the answer.

The second part is really easy. It took 37 mins to decay half way. meaning to decay another half of 50% which equals 25% it will take an additional 37 mins!

2)

omega = sqrt(k/m)

==> k = omega^2*m

= (2pi*f)^2*m = 239.04 N/m

3)

Using T = 2(m/k) for loaded spring in the form .. T² = 4²(m/k)

For chair unoccupied (m) .. .. .. . 0.842² = 4²(m/606)
For chair + astronaut (m+M) .. .. 2.280² = 4²([m+M]/606)

Subtracting .. (2.28² - 0.842²) = (4²/606)([m+M] - m)

4.489 = 0.065 x M .. .. .. M = 4.489 / 0.065 .. .. .. M = 69.06 kg

4)

Since, F =-kx
x max = F/k
k =m.2 = m.4.2f2
=0.10x10^-3 x4x3.14x3.14x10^6x10^6
=1.25x10^9
Amplitude,
A = x max = F/k = 38000/1.25x10^9
= 3.04x 10^-5m
Max speed,
vmax = .A = 2.f.A
= 6.282 x10^6x3.04x10^-5
=191 m/s

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