Question

The equation of a transverse wave travelling along a coil spring is y=4 sin π(0.010x-2t) where y and x are in cm and t in s. find the wavelength, initial phase at the origin ,speed and frequency on the wave

Answer 1

Answer:

tan2xtan(x+a)tan(x−a)dx

To evaluate this type of integrals, which is given in three multiples of tan together such that one angle is equals to sum of other two angles, we solve as follow.

From given statement, we have

\rm \: 2x = (x + a) + (x - a)2x=(x+a)+(x−a)

So,

\rm \: tan2x = tan[(x + a) + (x - a)]tan2x=tan[(x+a)+(x−a)]

We know,

\begin{gathered}\boxed{\tt{ tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany} \: }} \\ \end{gathered}

tan(x+y)=

1−tanxtany

tanx+tany

So, using this identity, we get

\rm \: tan2x = \dfrac{tan(x + a) + tan(x - a)}{1 - tan(x + a) \: tan(x - a)} tan2x=

1−tan(x+a)tan(x−a)

tan(x+a)+tan(x−a)

\rm \: tan2x - tan2x \: tan(x + a) \: tan(x - a) = tan(x + a) + tan(x - a)tan2x−tan2xtan(x+a)tan(x−a)=tan(x+a)+tan(x−a)

\rm\implies \: tan2x \: tan(x + a) \: tan(x - a) =tan2x - tan(x + a) - tan(x - a)⟹tan2xtan(x+a)tan(x−a)=tan2x−tan(x+a)−tan(x−a)

So,

\displaystyle\int\rm tan2x \: tan(x + a) \: tan(x - a) \: dx∫tan2xtan(x+a)tan(x−a)dx

can be rewritten as

\rm \: = \: \displaystyle\int\rm \: [ tan2x - \: tan(x + a) - \: tan(x - a)] \: dx=∫[tan2x−tan(x+a)−tan(x−a)]dx

We know,

\begin{gathered}\boxed{\tt{ \displaystyle\int\rm tanx \: = \: log |secx| + c \: }} \\ \end{gathered}

∫tanx=log∣secx∣+c

So, using this result, we get

\rm \: = \: \dfrac{1}{2}log |sec2x| - log |sec(x + a)| - log |sec(x - a)| + c=

2

1

log∣sec2x∣−log∣sec(x+a)∣−log∣sec(x−a

Explanation:

The equation of a transverse wave travelling along a coil spring is y=4 sin π(0.010x-2t) where y and x are in cm and t in s. find the wavelength, initial phase at the origin ,speed and frequency on the wave