Chapter 15, Problem 66A

To determine

To find:The new frequency of the plastic tube.

Answer to Problem 66A

The new frequency is $4.0\times {10}^2\text{ Hz}$ .

Explanation of Solution

Given data:

The previous frequency of the plastic tube when it was swung slower is $\left(f_1\right)\text{=2}\text{.0}\times {\text{10}}^2\text{Hz}$ .

Formula used:

  $f_2=2f_1$

Calculation:

  $\begin{array}{l}{f}_2=2f_1\\ =\left(2\right)2.0\times {10}^2\text{Hz}\\ =4.0\times {10}^{{}^2}\text{ Hz}\end{array}$

Conclusion:

The new frequency is $\text{4}\text{.0}\times {\text{10}}^2\text{ Hz}$ .