Sin wt- 1 - 2 - Find the value of t -

Solution we have, #160; sin(wt) = 1√(2) ⇒ wt = sin^ 1(1√(2)) = π4 ⇒ t = π4w As w (angler frequency) = 2πT(time period) ⇒ t = π (T)4(2π ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Property 2

sin wt= 1 /√...

Question

$s i n w t = \frac{1}{\sqrt{2}}$ , Find the value of t ?

Open in App

Solution

Solution we have,
$s i n (w t) = \frac{1}{\sqrt{2}}$
$\Rightarrow w t = s i n^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}$
$\Rightarrow t = \frac{π}{4 w}$
As w (angler frequency) $= \frac{2 π}{T (t i m e p e r i o d)}$
$\Rightarrow t = \frac{π (T)}{4 (2 π)} = \frac{T}{4 \times 2} = \frac{T}{8}$

Suggest Corrections

Similar questions

If vectors A^-->= cos wt i^{^} + sin wt j^{^} and B^-->= cos wt /2^{^}i + sin wt/2 j^{^}are functions of time, then value of t at which they are orthogonal to each other is:

Q. If vectors

\to A = cos w t^i + sin w t^j

and

\to B = cos w t / 2^i + sin w t / 2^j

are functions of time , then the value of

t

at which they are orthogonal to each other is

2 \int \frac{t d t}{t^{4} + 1} = \frac{1}{\sqrt{m}} t a n^{- 1} t^{2}

.Find the value of

m

(a) T_n = n find (i) T_n+1 (ii) T _n−1

(b) If T_n = n² − 1 find (i) T_n−2 (ii) T_n+1

(d) In a sequence T_n= 5 − 3n find (i) T_n+1 (ii) T_n+2

Q. If

s = \frac{t^{3}}{3} - \frac{1}{2} t^{2} - 6 t + 5

, find the value of

\frac{d^{2} s}{d t^{2}}

at t = 1.

Join BYJU'S Learning Program

sinwt=1√2, Find the value of t ?

Solution we have, sin(wt)=1√2⇒wt=sin−1(1√2)=π4⇒t=π4wAs w (angler frequency) =2πT(time period)⇒t=π(T)4(2π)=T4×2=T8

$s i n w t = \frac{1}{\sqrt{2}}$ , Find the value of t ?

Solution we have,
$s i n (w t) = \frac{1}{\sqrt{2}}$
$\Rightarrow w t = s i n^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4}$
$\Rightarrow t = \frac{π}{4 w}$
As w (angler frequency) $= \frac{2 π}{T (t i m e p e r i o d)}$
$\Rightarrow t = \frac{π (T)}{4 (2 π)} = \frac{T}{4 \times 2} = \frac{T}{8}$