Austin N. answered • 11/06/19
Experienced Tutoring in College Physics, SAT Math, and Regents Physics
First of all, recall that strings with standing wave patterns possess a frequency-wavelength-speed relationship. Specifically, the speed of a wave in terms of its wavelength and frequency is given by v=ƒλ where ƒ is the frequency and λ is the wavelength. We can also express the speed of a wave on a string by the tension on the string and its linear mass density. This relationship forms the equation v=√T/μ where T is the tension and μ is the linear mass density, or mass per unit length. What we want to do is use these relationships, along with the fact that the period can be found by t=1/ƒ, to determine the period of the oscillation.
Begin by noticing that the snapshot of the standing transverse wave is depicted as a sine curve going through two full cycles. Because of this, we say that the length of the string L is equivalent to two wavelengths. Therefore, L=2λ. Rearranging, we see that λ=L/2.
As mentioned above, the speed of a standing tranverse wave on a string is given by v=√T/μ. Since μ is the mass per unit length, we write μ=m/L where m stands for the total mass of the string. Substituting this relation into the speed equation we yield, v=√TL/m. Before we plug in the numbers, note that for dimensional reasons the mass must be converted from grams to kilograms. Doing so will give a value of m=0.00967kg. When we use this value, along with the given values for tension and length, we find v=31.7695945 meters per second.
Now, we know that v=ƒλ, which means that ƒ=v/λ. Because the period of the oscillation can be written as t=1/ƒ, we may also write t=λ/v. Remembering that λ=L/2, we can see that λ=0.61 meters. Now that we have both v and λ we can plug them into our equation for the period and yield t=λ/v=0.019200749 seconds.
