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Question

A simple harmonic motion is represented by y=5(sin3πt+3cos3πt)cm The amplitude and time period of the motion are :
  1. 10cm,23s
  2. 10cm,32s
  3. 5cm,23s
  4. 5cm,32s

A
10cm,23s
B
5cm,23s
C
10cm,32s
D
5cm,32s
Solution
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y=5(sin3πt+3cos3πt)
y=5×2(12sin3πt+32cos3πt)
y=10(cosπ3sin3πtsinπ3cos3πt)
y=10sin(3πt+π3)
Comparing it with y=Asin(ωt+ϕ)
A=10 i.e., amplitude is 10cm
T=2πω=2π3π=23 sec. i.e, time period is 23 sec.

1208659_1400738_ans_a47cc12ed1d84815851f44187b9d76bc.jpg

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