Question

The natural frequency of a string under constant tension varies inversely as its length. If a string 40 cm long vibrates 680 times per second, what length must the string be to vibrate 850 times per second under the same tension?

Answer 1

As the frequency varies inversely as the length of string:

$$f = \frac {K}{L}; L = \frac {K}{f}$$

Were:

$f =$ frequency, $L =$ long of string;

Let's find a coefficient of proportionality $K$:

$$40 = \frac {K}{680}; K = 40 * 680 = 27200;$$ $$L = \frac {K}{f} = \frac {27200}{850} = 32 \text{cm}.$$

Answer: 32 cm.