. The number of vibrations per second made when a stretched wire is plucked varies directly as the square root of the tension in the wire and inversely as the length.
If a 1.23 m long wire will vibrate 325 times each second when the tension is 115 N, determine the number of vibrations per second if the wire is shortened to 0.95 m and the tension is decreased to 92.0 N.
frequency = k √t/L
given: 325 = k√115/1.23
k = 1.23(325)/√115
for 2nd case
frequency = (1.23(325)/√115) √92/.95
= .....
you do the button-pushing
To solve this problem, we can use the concept of direct and inverse variation. Let's break down the given information and use it to find the number of vibrations per second in the second scenario.
Given information:
Length of wire (first scenario): 1.23 m
Number of vibrations per second (first scenario): 325
Tension in wire (first scenario): 115 N
We are asked to determine the number of vibrations per second in the second scenario:
Length of wire (second scenario): 0.95 m
Tension in wire (second scenario): 92.0 N
Using the given information, we can establish the following proportion:
(Length1)/(Length2) = (Tension2)/(Tension1) * (vibrations1)/(vibrations2)
Substituting the given values:
1.23 / 0.95 = 92.0 / 115 * 325 / vibrations2
Simplifying the proportion:
1.23 / 0.95 = 29900 / vibrations2
To find the number of vibrations per second in the second scenario, we need to rearrange the equation to solve for vibrations2:
vibrations2 = 29900 / (1.23 / 0.95)
Simplifying further:
vibrations2 = 29900 * (0.95 / 1.23)
vibrations2 ≈ 23088.62
Therefore, the number of vibrations per second when the wire is shortened to 0.95 m and the tension is decreased to 92.0 N is approximately 23088.62.