I failed trying to isolate variable t.

Wolfram Alpha did it for me but not showing the successive steps. Tried using trigonometric identities but no joy.

Could anyone detail the steps? Thanks.

 

I can't make out your equation. Is the left-hand side a theta with a subscripted letter (which letter? h? p?)? Is it a ratio of theta and something else (r?)? Is that a function named "sen"?

Assuming that this is related to your prior pendulum thread, I'm going to guess that you equation is

\(\theta_h \; = \; \theta_m \sin \( {\sqrt{\frac{x}{m}} t }\)\)

So divide both sides by theta_m

\(\sin \( {\sqrt{\frac{x}{m}} t }\) \; = \; \frac{\theta_h}{\theta_m}\)

Take the arcsin of both sides (which I'll write using the exponential inverse convention)

\(\sqrt{\frac{x}{m}} t \; = \; \sin^{-1} \( \frac{\theta_h}{\theta_m} \)\)

Now divide both sides by sqrt(x/m)

\(t \; = \; \sqrt{\frac{m}{x}} \sin^{-1} \( \frac{\theta_h}{\theta_m} \)\)

I don't know what 'm' and 'x' are, but the ratio better have units of time-squared.

 

I can't make out your equation. Is the left-hand side a theta with a subscripted letter (which letter? h? p?)? Is it a ratio of theta and something else (r?)? Is that a function named "sen"?

Assuming that this is related to your prior pendulum thread, I'm going to guess that you equation is

\(\theta_h \; = \; \theta_m \sin \( {\sqrt{\frac{x}{m}} t }\)\)

So divide both sides by theta_m

\(\sin \( {\sqrt{\frac{x}{m}} t }\) \; = \; \frac{\theta_h}{\theta_m}\)

Take the arcsin of both sides (which I'll write using the exponential inverse convention)

\(\sqrt{\frac{x}{m}} t \; = \; \sin^{-1} \( \frac{\theta_h}{\theta_m} \)\)

Now divide both sides by sqrt(x/m)

\(t \; = \; \sqrt{\frac{m}{x}} \sin^{-1} \( \frac{\theta_h}{\theta_m} \)\)

I don't know what 'm' and 'x' are, but the ratio better have units of time-squared.

@WBahn Gracias for your detailed reply.

Sure it is related to the pendulum thread.
Sen (seno) is my native version of sin.

Shame to me when I saw you taking arcsin of both sides...! I failed there maybe because in my mindset, radians is a concept isolated from the rest. Hard to get used to it.

Sorry for not replying earlier. Today is my fifth day in black out at home. Using my mobile in my usual cafeteria to access AAC.

Thanks again.

 

Glad to help -- it's real easy to get a mental block and not see what is staring you in the face. Worse, the longer it stares you in the face, the easier it is to miss it because you've already established in your mind that it's not there to be seen.

Five days without power -- yikes. Weather related? Hope you get power back soon.