Two Mechanics Questions 83, 84

[yxgao]

There is a diagram that accompanies it so if the explanation isn't clear you can refer to the sample test posted on

the GRE.org website.

83 on GRE. Consider a particle moving without friction on a rippled surface, as shown above. Gravity acts down in the negative h direction. The elevation h(x) of the

surface is given by h(x) = d cos[kx]. If the particle starts at x=0 with a speed v in the x direction, for what values of v will the particle stay on the

surface at all times?

The answer is v <= Sqrt[g/(k^2*d)]
Why, and what concepts are involved here?

84. Two pendulums are attached to a massless spring, as shown above. The arms of the pendulums are of identical lengths l, but the pendulum balls have unequal

masses m1 and m2. The initial distance between the masses is the equilibrium length of the spring, which has spring constant K. What is the highest normal

mode frequency of the system?
A. Sqrt[g/l]
B. Sqrt[K/(m1+m2)]
C. Sqrt[K/m1 + K/m2]
D. Sqrt[g/l + K/m1 + K/m2]
E. Sqrt[2g/l + K/(m1+m2)]

The answer is:
D. Sqrt[g/l + K/m1 + K/m2]
My question is why, and how do you know that this frequency is the highest?

 

[arcnets]

Originally posted by yxgao
83 on GRE.
The answer is v <= Sqrt[g/(k^2*d)]
Why, and what concepts are involved here?

If the speed is too high, the particle will lift off because of the centrifugal force. You can calculate the speed by the conservation of energy. You can find the centrifugal force if you know the radius of curvature of the path. Remember the lift-off can only happen where the path is convex (curved downward).

84. Two pendulums...
My question is why, and how do you know that this frequency is the highest?

I think it's obvious that the motion depends on both the values of g and K, because gravity acts against/with the spring force. So only D or E can be correct. Now imagine K was zero, then surely you have two independent pendulums with period sqrt(g/l) each. Thus, D must be correct.

 

[M98Ranger]

For question 83, the particle is moving on a rippled surface, where the elevation of the surface is given by h(x) = d cos[kx]. The particle is also acted upon by gravity in the negative h direction. In order for the particle to stay on the surface at all times, the force of gravity must be balanced by the normal force from the surface. This means that the acceleration in the h direction must be zero. Using Newton's second law, F=ma, we can set up the following equation:

ma = -mg + N

Where m is the mass of the particle, g is the acceleration due to gravity, and N is the normal force. Since the particle is not moving in the h direction, a=0 and we can solve for N:

N = mg

We can also express the normal force in terms of the elevation of the surface, h(x):

N = m(d cos[kx])

Setting these two expressions for N equal to each other, we can solve for the maximum speed v that will keep the particle on the surface at all times:

mg = m(d cos[kx])

Simplifying and solving for v, we get:

v <= Sqrt[g/(k^2*d)]

This means that for values of v that are less than or equal to this maximum speed, the particle will stay on the surface at all times. The concepts involved here are Newton's second law, forces, and kinematics.

For question 84, we can use the formula for the frequency of a simple harmonic oscillator:

f = 1/2pi * Sqrt(K/m)

Where K is the spring constant and m is the mass. In this case, we have two pendulums with different masses, but they are attached to the same spring. This means that the effective mass of the system will be the sum of the individual masses, m1 + m2. The equilibrium position of the system is when the spring is at its natural length, so we can use this as our reference point for calculating the frequency. This means that the frequency will be determined by the effective mass of the system and the spring constant. Using the formula above, we get:

f = 1/2pi * Sqrt(K/(m1+m2))

However, this is not the highest frequency that the system can have. In order to find the highest frequency, we need to consider the other forces acting on the system. The pendul