A stretched wire 75cm long emits a fundamental frequency of 300Hz.By how much the length be changed so that it emits a fundamental note of 450Hz?
The frequency of the fundamental mode of oscillation is inversely proportional to length of string, so we can write L2/L1 = f1/f2 where L and f are length and frequency respectively, and subscripts 1 & 2 refer to 2 different cases of f & L.
Using your data we get L2 = f1*L1/f2 = 50cm Change of length = -25cm.
Alternatively you could argue that the frequency rises by a factor 3/2, so the length falls by a factor 2/3 to get the new length of 50cm...
Qn.1.Ans.25cm
To find the change in length needed to emit a fundamental note of 450 Hz, we can use the formula for the frequency of a stretched wire:
f = (1/2L) * √(T/μ)
where:
f is the frequency,
L is the length of the wire,
T is the tension in the wire, and
μ is the linear mass density of the wire.
Let L1 be the initial length of the wire (75 cm) and f1 be the initial frequency (300 Hz). Let L2 be the new length we are trying to find, and f2 be the desired frequency (450 Hz).
We can set up the following equation:
f1 = (1/2L1) * √(T/μ)
f2 = (1/2L2) * √(T/μ)
To eliminate T and μ, we can divide these equations:
f1 / f2 = (L2 / L1)
Now we can solve for L2:
L2 = (f2 / f1) * L1
= (450 Hz / 300 Hz) * 75 cm
L2 = 1.5 * 75 cm
L2 = 112.5 cm
Therefore, the length needs to be changed to 112.5 cm in order for the wire to emit a fundamental note of 450 Hz.
To determine how much the length of the wire must be changed in order to emit a fundamental frequency of 450Hz, we need to use the formula:
f = (1/2L) * sqrt(T/m)
where:
f = fundamental frequency
L = length of the wire
T = tension in the wire
m = linear mass density of the wire
In this case, we are given that the wire is initially 75cm long and emits a fundamental frequency of 300Hz. We'll denote this length as L1 and frequency as f1. We want to find the change in length (denoted as ΔL) that will result in the desired frequency of 450Hz, denoted as f2.
Let's first calculate the initial tension in the wire, T1, for the given frequency f1 using the formula:
T1 = (4L1^2 * f1^2 * m) / π^2
To find the desired tension T2 at frequency f2, we rearrange the equation as follows:
T2 = (π^2 * T1 * f2^2) / (4L1^2 * m)
Now, we know that the tension T2 is the same as T1 because the wire is unchanged except for its length. Rearranging the equation to solve for the new length L2, we have:
L2 = sqrt((π^2 * L1^2 * f2^2 * m) / (f1^2 * m))
Plugging in the given values:
L1 = 75cm
f1 = 300Hz
f2 = 450Hz
And assuming a constant linear mass density of the wire:
m = 1 (since it's not given)
We can now calculate the new length L2:
L2 = sqrt((π^2 * 75^2 * 450^2) / (300^2))
Simplifying the expression:
L2 = sqrt(π^2 * 75^2 * 1.5)
L2 ≈ 75 * 1.2255
L2 ≈ 91.9125 cm
Therefore, the length of the wire must be changed to approximately 91.9125 cm in order to emit a fundamental frequency of 450Hz.