Sprint 8:36 PM The graph below shows data taken on the position of a small massi

Question

Sprint 8:36 PM The graph below shows data taken on the position of a small massive cart attached to a spring as shown in the figure at the right. Find the period (T), frequency (n, and angular frequency () of the oscillation. 1/s = 1354 rad/s B. If the spring constant is k 2.4 N/m, can you find the mass of the cart? If you can do it, give the answer here. If you can't, put 0 The simple mass on a spring is the basis for our description of almost any oscillation. Lets consider the vibrational states of a simple diatomic molecule, say N We will model it as two identical masses connected by a spring as shown in the figure at the top right. This is a little different from the cart on a spring since there is no wall - both of the atoms are moving. But if we only look at oscillations of the molecule where the center of mass is not moving, the atoms are moving equally and oppositely. When one is going left, the other goes an equal amount to the right and vice versa. This is like two carts attached to a wall and oscillating equally and oppositely as shown in the lower picture. The only difference is the way we treat the spring. When each atom moves a distance x, the spring between the atoms actually stretches a distance 2x, so the force on each atom is 2kx. We can therefore model the motion of one of the nitrogen atoms as a single cart on a spring if we replace k by 2k. C. If the angular frequency of oscillation of N2 is found to be 4.5 x 1018 rad/s, and the mass of a single nitrogen atom is 2.3 x 10-26 kg, find the effective spring constant between the two atoms. (Give your answer in N/nm, since nanometers is a more appropriate scale for atoms than meters.) A mass hanging from a vertical spring is somewhat more complicated than a mass attached to a horizontal spring because the gravitational force acts along the direction of motion. Therefore, the restoring force of the oscillations is not provided

Explanation / Answer

Write the relation between angular frequency and spring constant in simple harmonic motion -

w = sqrt[k/m]

here we have to consider 2k in stead of k for the spring constant.

so the expression becomes -

w = sqrt[2k/m]

=> w^2 = 2k/m

=> k = m*w^2/2 = [2.3x10^-26x(4.5x10^18)^2]/2 = 23.29 x 10^10 N/m = 23.29 x 10 N/nm = 232.9 N/nm.

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