78.5k views
0 votes
What is the Laplace inverse of s/(L^2s^2+n^2π^2)?

a. sin(nπt)

b. sin(nπt)/L

c. cos(nπt)

d. cos(nπt)/L

1 Answer

3 votes

Final answer:

The inverse Laplace transform of s/(L^2s^2+n^2π^2) is b. sin(nπt)/L, which resembles the Laplace transform of a sine function with a specific frequency related to n and L.

Step-by-step explanation:

The inverse Laplace transform of the function s/(L^2s^2+n^2π^2) is the function of time, t, which when transformed into the s-domain (Laplace domain) will yield the original function.

To find the inverse Laplace transform, one can look for a known correspondence in a table of Laplace transforms or apply the inverse transformation techniques.

The given function resembles the transform of a sine function, and specifically, it has the form of the transform for sin(bt), where b = nπ/L.

The Laplace transform of sin(bt) is b/(s^2+b^2).

By comparing this with our original function, we can conclude that the correct inverse transform will be sin(nπt/L), which corresponds to option (b) sin(nπt)/L in the question.

User Smitha
by
7.6k points