TS EAMCET 20-July-2022 Shift 1 Solved Paper

Let displacement of particle in SHM is
x=A‌cos(ωt+φ)
Then, it's velocity is
v=‌

dx

dt

=−Aωsin‌(ωt+φ)
Now, at t=2 s,x=0
and ω=‌

2π

T

, where T=16 s
So, 0=A‌cos(‌

2π

16

×2+φ)
‌⇒‌‌‌‌cos(‌

π

4

+φ)=0
‌⇒‌‌‌

π

4

+φ=‌

π

2

‌⇒‌‌φ=‌

π

4

Also, velocity is 4m∕ s at t=4 s. So, v=−Aω(sin‌(ωt+φ)) gives,
⇒4=−A(‌

2π

16

)⋅sin‌(‌

2π

16

×4+‌

π

4

)
⇒4=−A×‌

2π

16

×sin‌(‌

π

2

+‌

π

4

)
⇒4=−A×‌

2π

16

×cos‌

π

4

.
⇒A=‌

−32√2

π

or magnitude of amplitude is ‌

32√2

π

units.