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A particle executing S.H.M. in a straight line has velocities 8, 7, 4 at three points distant one foot from each other. The maximum velocity of the particle will be
- a)√28
- b)√32
- c)√56
- d)√65
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A particle executing S.H.M. in a straight line has velocities 8, 7, 4 ...
We know that 
∴
64 - 49 = 15 = ω2 (2 x + 1)
64 - 16 = 48 = ω2 (4 x + 4)
∴
further,
ω = 3
Maximum velocity = ωa = 3 x [√65/3] = √65 Ans
∴
64 - 49 = 15 = ω2 (2 x + 1)
64 - 16 = 48 = ω2 (4 x + 4)
∴
further,
ω = 3
Maximum velocity = ωa = 3 x [√65/3] = √65 Ans
Most Upvoted Answer
A particle executing S.H.M. in a straight line has velocities 8, 7, 4 ...
To find the maximum velocity of the particle, we need to determine the amplitude of the SHM.
Let's assume the displacement of the particle from its equilibrium position at the three given points is x1, x2, and x3 respectively.
Then, we can use the formula for velocity in SHM:
v = ±ω√(A² - x²)
where v is the velocity, ω is the angular frequency, A is the amplitude, and x is the displacement from the equilibrium position. The ± sign is used to indicate the direction of motion.
Using the given velocities, we can write three equations:
8 = ±ω√(A² - x1²)
7 = ±ω√(A² - x2²)
4 = ±ω√(A² - x3²)
Squaring all three equations and subtracting the second from the first and the third from the second, we get:
1 = 4x1² - 4x2² + x3²
9 = 4x2² - x1² - 4x3²
Solving these two equations simultaneously, we get:
x1 = 0.5 ft
x2 = 1.0 ft
x3 = 1.5 ft
Substituting these values in any of the three original equations, we can solve for the amplitude:
8 = ±ω√(A² - 0.25)
A² - 0.25 = 64/ω²
A² = 64/ω² + 0.25
Taking the derivative of A² with respect to ω and setting it to zero, we can find the value of ω that maximizes A²:
d(A²)/dω = -128/ω³ = 0
ω = (128)^(1/3) ≈ 5.039
Substituting this value of ω in the equation for A², we get:
A² ≈ 16.04
Taking the square root, we get:
A ≈ 4.007 ft
Therefore, the maximum velocity of the particle is:
vmax = ωA = (128)^(1/3) × 4.007 ≈ 20.19 ft/s
Let's assume the displacement of the particle from its equilibrium position at the three given points is x1, x2, and x3 respectively.
Then, we can use the formula for velocity in SHM:
v = ±ω√(A² - x²)
where v is the velocity, ω is the angular frequency, A is the amplitude, and x is the displacement from the equilibrium position. The ± sign is used to indicate the direction of motion.
Using the given velocities, we can write three equations:
8 = ±ω√(A² - x1²)
7 = ±ω√(A² - x2²)
4 = ±ω√(A² - x3²)
Squaring all three equations and subtracting the second from the first and the third from the second, we get:
1 = 4x1² - 4x2² + x3²
9 = 4x2² - x1² - 4x3²
Solving these two equations simultaneously, we get:
x1 = 0.5 ft
x2 = 1.0 ft
x3 = 1.5 ft
Substituting these values in any of the three original equations, we can solve for the amplitude:
8 = ±ω√(A² - 0.25)
A² - 0.25 = 64/ω²
A² = 64/ω² + 0.25
Taking the derivative of A² with respect to ω and setting it to zero, we can find the value of ω that maximizes A²:
d(A²)/dω = -128/ω³ = 0
ω = (128)^(1/3) ≈ 5.039
Substituting this value of ω in the equation for A², we get:
A² ≈ 16.04
Taking the square root, we get:
A ≈ 4.007 ft
Therefore, the maximum velocity of the particle is:
vmax = ωA = (128)^(1/3) × 4.007 ≈ 20.19 ft/s
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A particle executing S.H.M. in a straight line has velocities 8, 7, 4 at three points distant one foot from each other. The maximum velocity of the particle will bea)√28b)√32c)√56d)√65Correct answer is option 'D'. Can you explain this answer?
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