Maths Problem - Need all the help I can get!

Sorry for X-Posting but needing help quickly, hopefully someone can assist. The problem: Two voltages v1 and v2 are to be summed using an analogue circuit.
v1 = 5 cos (5 pi t + (pi / 6)) v2 = 10 sin (5 pi t - (pi / 6)) Calculate the time at which the sum is first equal to 4 volts. I have added the two voltages and left the result in R = sin (wt + a) and was calculated in radians w = omega a = alpha My answer to the sum is 11.18 sin (5 pi t + 1.63) Now, how do I find the time when the voltage is first equal to 4 volts. Appreciate any help. TIA
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
BIGEYE wrote:

Solve the equation 11.18 sin (5 pi t + 1.63) = 4. There are an infinite number of solutions, but you can get the smallest positive one using the inverse sine function. -- Christopher Heckman
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
On Sun, 9 Jan 2005, BIGEYE wrote:

Huh, you cannot finish? Solve     11.18 sin(5pi.t + 1.63) = 4 for t. It will have a number of values. I'd consider the smallest positive value for t, to be when the voltage is first 4 volts.
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload
Your value of 11.18 is incorrect by my reconning. I make it v = 6.196 sin(5pi t - 0.1077) From: v1 = 4.33cos(5pi t) - 2.5sin(5pi t) and v2 = 8.66sin(5pi t) - 5cos(5pi t) v1 +v2 = v = 6.16sin(5pi t) - 0.67cos(5pi t) which gives: v = 6.196sin(5pi t - 0.1077) when v = 4Volts we have 4/6.196 = sin(5pi t - 0.1077) = 0.6456 therefore: arcsin(0.6456) = 5pi t - 0.1077 = 0.702 solving for t gives 0.0515 seconds If you ave access to Matlab (or similar) you can verify my calculations for yourself. Hope this helps John Ferguson Windsor, England
Add pictures here
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Add image file
Upload

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here. All logos and trade names are the property of their respective owners.