13. An object is hung on a spring, and the frequency of oscilation of the system, f, is measured. The object, a second identical object, and the spring are carried to space in the Space Shuttle. The two objects are attached to the ends of the spring, and the system is taken into space on a space walk. The spring is extended, and the system is released to oscillate while floating in space. The coils of the spring do not bump into one another. What is the frequency of oscillation for this system, in terms of f?
The question again.
Consider:
When the spring w/ the 2 objects on opposite ends is set into oscillation is space the coil at the exact center does not move. Therefore, we can imagine clamping the center coil in place without affecting the motion. If we do this, we have 2 separate oscillating systems, one on each side of the clamp...
Originally posted by Ramnedhmmm.
You can consider anything to logically solve the problem. But it is not necessary to calculate anything. Takes perhaps a bit of creativity.
I'll start you off, since everyone is lost.
When the spring w/ the 2 objects on opposite ends is set into oscillation is space the coil at the exact center does not move. Therefore, we can imagine clamping the cente ...[text shortened]... motion. If we do this, we have 2 separate oscillating systems, one on each side of the clamp...
This is my last shot. Each subsystem is a replica of the previous system only that this time the spring has half the size. But that, as fas as I remember, doesn't affect the spring constant of the material. So both subsystems osccilate with the same frequency. Since we are assuming Hooke's Law is valid this system is linear and in that case the principle of sobreposition is valid so we can say that the new frequency is 2f.
great clarification question, peteyjames, but in prior questions, when we gave a partial correct answer, or our reasoning was getting close, he gave us some feedback telling us where we were getting warm, or going far astray, so i was assuming that f, 2f and 4 f were all wrong, regardless of the reasons given.
Originally posted by PeteyJamesI'm really not paying attention to the explantations when I see that the answer is wrong. The person who has been closest is adamwarlock - his first guess.
so can we confirm that the answer is not f, 2f or 4f? or is it the reasons that are incorrect?
I sent a PM to AThousandYoung, mtthw, and sven1000 to see if they'd post.
The frequency of oscillation, as adam warlock said early on, is proportional to sqrt(k/m). Thus it depends only on the spring constant and the inertial mass attached to the spring. The inertial mass doesn't change between Earth and space. The spring constant of the entire spring doesn't change either. So what gives?
In the second part of the question, we have two masses, each stretching the spring. This means, instead of a Hooke's Law equation F= -k*x, we have F= -k*(2*x), where k is the spring constant again, and x the displacement from the equilibrium position. Thus the equivalent spring constant for each half is twice that of the first part of the problem. When used in f= sqrt(k/m) (the 2*pi isn't important in a ratio answer) we see that in space, the frequency of oscillation is HIGHER by a factor of sqrt(2) for both objects.
So the reason for the frequency changing is due to the two objects attached to the spring, versus one object initially. The result would be the same in space or if the system were laid on a frictionless table on Earth.
Originally posted by sven1000Very well done! That is correct.
The frequency of oscillation, as adam warlock said early on, is proportional to sqrt(k/m). Thus it depends only on the spring constant and the inertial mass attached to the spring. The inertial mass doesn't change between Earth and space. The spring constant of the entire spring doesn't change either. So what gives?
In the second part of the que sult would be the same in space or if the system were laid on a frictionless table on Earth.
Here is how I thought it through.
When the spring w/ the 2 objects on opposite ends is set into oscillation in space, the coil at the exact enter of the spring doesn't move. So we can imagine clamping the center coil in place w/o affecting the motion. If we do this, we have two separate oscillating systems, one on each side of the clamp. The half spring on each side of the clamp has twice the spring constant of the full spring, as shown by this argument: The force exerted by a spring is proportional to the separation of the coils as the spring is extended. Imagine that we extend a spring by a given distance and measure the distance between coils...and cut the spring in half. If one of the half-springs is now extended by the same distance, the coils will be twice as far apart as they were in the complete spring. Thus, our clamped system of two objects on two half springs will vibrate with a frequency that is higher than f by a factor of r(2).
Ok. That was complicated...
I'm too tired to think of the next question. It's going to be on sound. I'll post it tomorrow. Congratulations sven1000 🙂
Wind instruments make sound because of the resonating sound waves within the instrument. The lengths of the resonance waves vary depending on the speed of sound. The speed of sound varies with air temperature. Therefore the length of the resonance waves varies with the air temperature.
Also to a minor degree, the change of air temperature probably changes pressure and humidity, both of which change the speed of sound.
To an even more minor degree, changing the temperature of the instruments changes their resonance characteristics.
Originally posted by RamnedPitch is frequency right? Well the temperature is proportional to pressure by the Ideal Gas Law, and sound waves are pressure waves. The temperature changes the pressure which changes the sound waves and causes different perceived pitch.
Not exactly answering the question: How does it affect pitch; what cause by temperature effects pitch?
higher temperatures cause wind instruments (esspecially the metal ones) to expand, making them slightly longer, and changing the pitch
you tune an instrument by slidint the mouthpiece further in/out. Temperature essentially does this by expanding the instrument itself rather than sliding the mouthpiece.