 Table mass and the strict displacement bound

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2/section-b/b6/table-mass-and-the-strict-displacement-bound

The four supporting springs act in parallel, so their total spring constant is

k = 4.00 \times 1 0^{4} N m^{- 1}

ω = 2 π (10) = 62.83 s^{- 1}

∣ x_{0} ∣ ≃ \frac{F _{0}}{m ω ^{2}}, m_{estimate} = \frac{5}{( 2 π 10 ) ^{2} ( 1 0 ^{- 6} )} = 1.267 \times 1 0^{3} kg .

At this mass,

f_{0} = k / m / (2 π) = 0.894 Hz ≪ 10 Hz

, and

Q = mk / b ≃ 14.2

, so the assumptions are consistent. The exact forced amplitude is

∣ x_{0} ∣ = \frac{5}{( 4.00 \times 1 0 ^{4} - m ω ^{2} ) ^{2} + ( 500 ω ) ^{2}} = 1.008 μ m .

It is indeed close to

1 μ m

, but it is slightly above the strict requested bound. Setting the exact expression to

1 0^{- 6} m

gives the high-mass threshold

m_{m i n} = \frac{k + ( F _{0} / x _{m a x} ) ^{2} - b ^{2} ω ^{2}}{ω ^{2}} = 1.277 \times 1 0^{3} kg .

Choose a mass strictly greater than this threshold for a strict inequality;

1.28 \times 1 0^{3} kg

is a suitable rounded choice. Simply rounding the high-frequency estimate down would not guarantee the bound.

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