The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes 2 cm, 1 m/s and 10 m/s² at a certain instant. Find the amplitude and the time period of the motion.
Answers
Answered by
60
x = 2 cm, v = 1 m/s = 100 cm/s., and a = 10 m/s². = 1000 cm/s².
Using the Formula,
a = -ω²x
∴ 1000 = -2 × ω²
ω² = -500
negative symbol was just showing the opposite direction. It does not have any significance here.
∴ ω = 10√5
T = 2π/10√5 = π/5√5
_____________________
Now,
v² = ω²(A² - x²)
∴ 10000 = 5 × 100(A² - 4)
∴ A² - 4 = 20
∴ A² = 24
A = ± 4.9 cm.
Hope it helps .
Answered by
0
Answer:
x = 2 cm, v = 1 m/s = 100 cm/s., and a = 10 m/s². = 1000 cm/s².
Using the Formula,
a = -ω²x
∴ 1000 = -2 × ω²
ω² = -500
negative symbol was just showing the opposite direction. It does not have any significance here.
∴ ω = 10√5
T = 2π/10√5 = π/5√5
_____________________
Now,
v² = ω²(A² - x²)
∴ 10000 = 5 × 100(A² - 4)
∴ A² - 4 = 20
∴ A² = 24
A = ± 4.9 cm.
Hope it helps .
Explanation:
Similar questions