find phase angle
x = 15 sin (wt + n/6) ;
x2 = 8 cos (wt + n/3)
Answers
Answered by
0
Answer:
First SHM is given as x
1
=asinwt+acoswt=a[sinwt+coswt]
∴ x
1
=
2
a[
2
1
sinwt+
2
1
coswt]
OR x
1
=
2
a[cos(π/4)sinwt+sinπ/4coswt]
⟹ x
1
=
2
asin(wt+π/4)
Second SHM is given as x
2
=asinwt+
3
a
coswt=a[sinwt+
3
1
coswt]
∴ x
2
=
3
2
a[
2
3
sinwt+
2
3
3
coswt]
OR x
2
=
3
2
a[
2
3
sinwt+
2
1
coswt]
OR x
2
=
3
2
a[cos(π/6)sinwt+sinπ/6coswt]
⟹ x
2
=
3
2
asin(wt+π/6)
Thus ratio of magnitude =
2a/
3
2
a
=
2
3
Also phase difference Δϕ=
4
π
−
6
π
=
12
π
Answered By
toppr
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