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Earthquake Damping Envelope (Civil & Structural Engineering) Solution

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Problem Overview

We are asked to evaluate the initial transient displacement limit of the tuned mass damper:

lim⁡t→0+t2sin⁡(50t)\lim_{t \to 0^+} t^2 \sin\left(\frac{50}{t}\right)
using the Squeeze Theorem.

Step-by-Step Mathematical Solution

Step 1: Establish the trigonometric range bounds

For any non-zero time t>0t > 0, the argument 50t\frac{50}{t} produces rapid oscillation, but the output of the sine function remains strictly bounded between −1-1 and 11:

−1≤sin⁡(50t)≤1-1 \le \sin\left(\frac{50}{t}\right) \le 1

Step 2: Multiply by the quadratic damping envelope t2t^2

Because t>0t > 0, t2t^2 is strictly positive (t2>0t^2 > 0). Multiplying all three parts of the inequality by t2t^2 preserves the inequality signs (see Solving Inequalities):

−t2≤t2sin⁡(50t)≤t2-t^2 \le t^2 \sin\left(\frac{50}{t}\right) \le t^2

Step 3: Evaluate the limits of the lower and upper bounding envelopes

Compute the limits of the outer boundary functions as t→0+t \to 0^+:

lim⁡t→0+(−t2)=−(0)2=0\lim_{t \to 0^+} \left(-t^2\right) = -(0)^2 = 0
lim⁡t→0+(t2)=(0)2=0\lim_{t \to 0^+} \left(t^2\right) = (0)^2 = 0

Step 4: Apply the Squeeze Theorem

Because −t2≤x(t)≤t2-t^2 \le x(t) \le t^2 and both bounding parabolas pinch together to 00 as t→0+t \to 0^+, by the Squeeze Theorem:

lim⁡t→0+x(t)=lim⁡t→0+t2sin⁡(50t)=0 cm\lim_{t \to 0^+} x(t) = \lim_{t \to 0^+} t^2 \sin\left(\frac{50}{t}\right) = 0\text{ cm}

Engineering Conclusion

Even though the damper experiences extreme vibrational frequency as t→0+t \to 0^+, the quadratic envelope t2t^2 successfully dampens the amplitude to zero. This mathematical proof verifies that the damper starts from a state of zero displacement without undergoing discontinuous initial impulse jumps that would snap mounting brackets or fracture high-rise structural anchor bolts.