When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12 Hz. When another object of mass m2 is hung on the spring along with m1, the frequency of the motion is 4 Hz. Find the ratio m2/m1 of the masses.
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Physics HELP!!!!!!!! - bobpursley, Wednesday, April 4, 2007 at 9:12pm
w= sqrt (k/m)
w^2= k/m
(w1/w2)^2=9 = (m1+m2)/m1
9= 1 + m2/m1
check my thinking.
Now, Bob why did u equate 9 to (m1 + m2)/m1?
cos one mass wass added to the other. Man, I wish i could break down questions like you do
The ratio of the hanging masses (m1+m2)/m1 is inversely proprtional to square of the ratio of the frequencies (which is 9 in this case). That follows from the equation
w= sqrt (k/m), and the fact that k is constant.
An object in static equilibrium has a coefficient of static friction as 0.110. If the normal force acting on the object is 95.0 newtons, what is the mass of the object?
the ans is
To find the ratio of the masses m2/m1, we can use the equation w^2 = k/m, where w is the angular frequency and k is the spring constant.
In the first scenario, when only the mass m1 is hung on the spring, the frequency is 12 Hz. The angular frequency can be calculated by using the formula w = 2πf, where f is the frequency. So, w1 = 2π(12) = 24π.
In the second scenario, when both masses m1 and m2 are hung on the spring, the frequency is 4 Hz. Again, calculating the angular frequency gives w2 = 2π(4) = 8π.
Now, we can use the equation (w1/w2)^2 = (m1 + m2)/m1. Plugging in the values, we get (24π/8π)^2 = (m1 + m2)/m1.
Simplifying, we have (3)^2 = (m1 + m2)/m1, which gives us 9 = 1 + m2/m1.
Finally, subtracting 1 from both sides of the equation, we get m2/m1 = 9 - 1 = 8.
Therefore, the ratio m2/m1 of the masses is 8.