Question
Suppose the kinetic energy of a body oscillating with amplitude $$A$$ and at a distance $$x$$ is given by
$$K = \frac{{Bx}}{{{x^2} + {A^2}}}$$
The dimensions of $$B$$ are the same as that of
A.
$$\frac{{{\text{work}}}}{{{\text{time}}}}$$
B.
$${\text{work}} \times {\text{distance}}$$
C.
$$\frac{{{\text{work}}}}{{{\text{distance}}}}$$
D.
$${\text{work}} \times {\text{time}}$$
Answer :
$${\text{work}} \times {\text{distance}}$$
Solution :
From $$K = \frac{{Bx}}{{{x^2} + {A^2}}} = \frac{{Bx}}{{{x^2}}} = \frac{B}{x}$$
$$\eqalign{
& \therefore B = K \times x = K.E. \times {\text{distance}} \cr
& = {\text{work}} \times {\text{distance}} \cr} $$