Solving for Natural Frequency in a Si-Based NEMS Resonator

[Cookiesm8]

NEMS Resonator

I need a hint to help with my homework.

The question:

"A Si-Based NEMS resonator has dimensions length=150nm, width=5nm and thickness=5nm and an effective spring constant of 100Nm^-1 and natural frequency of vibration of 250MHz"

I am looking for an equation linking the natural frequency to the effective spring constant, as for the first part of the question i have to show what would happen to them if the dimensions of the resonator were doubled.

Any advice, been trawling the net and my notes for a while and no joy.

 

[AvroArrow]

Hint: The equation linking the natural frequency and the effective spring constant is given by the equation f = 1/(2π√(k/m)), where f is the natural frequency, k is the effective spring constant, and m is the mass of the resonator.

 

[wais]

To solve for the natural frequency in a Si-Based NEMS resonator, you can use the equation:

f = 1/2π * √(k/m)

where f is the natural frequency, k is the effective spring constant, and m is the mass of the resonator.

To show what would happen to the natural frequency and effective spring constant if the dimensions of the resonator were doubled, you can use the equation:

k = 1/2 * E * (t/w)^3 * (l/w)

where E is the Young's modulus, t is the thickness, w is the width, and l is the length.

By doubling the dimensions, we can see that the thickness and length will both double, while the width remains the same. This means that the effective spring constant will increase by a factor of 8 (2^3), while the mass will increase by a factor of 4 (2*2). Plugging these values into the first equation, we can see that the natural frequency will increase by a factor of 2 (√(8/4) = 2).

Therefore, if the dimensions of the resonator were doubled, the natural frequency would increase by a factor of 2 and the effective spring constant would increase by a factor of 8.