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A particle of mass 250 g executes a simple harmonic motion under a periodic force F = (-25x)N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.
Correct answer is '40'. Can you explain this answer?
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Given data:
- Mass of the particle, m = 250 g = 0.25 kg
- Force acting on the particle, F = -25x N
- Maximum speed of the particle, vmax = 4 m/s
Step 1: Finding the angular frequency, ω
The force acting on the particle in simple harmonic motion is given by F = -kx, where k is the spring constant and x is the displacement from the equilibrium position.
Comparing this with the given force F = -25x N, we can equate the two expressions to find k.
-25x = -kx
Dividing both sides by x, we get:
-25 = -k
k = 25 N/m
The angular frequency, ω, is given by ω = √(k/m).
Substituting the values, we have:
ω = √(25/0.25) = √100 = 10 rad/s
Step 2: Finding the amplitude, A
The maximum speed of the particle, vmax, is related to the amplitude, A, and angular frequency, ω, by the equation:
vmax = Aω
Substituting the given values, we have:
4 = A * 10
Dividing both sides by 10, we get:
A = 4/10 = 0.4 m
Converting the amplitude from meters to centimeters:
Amplitude in cm = 0.4 * 100 = 40 cm
Therefore, the amplitude of the motion is 40 cm.
- Mass of the particle, m = 250 g = 0.25 kg
- Force acting on the particle, F = -25x N
- Maximum speed of the particle, vmax = 4 m/s
Step 1: Finding the angular frequency, ω
The force acting on the particle in simple harmonic motion is given by F = -kx, where k is the spring constant and x is the displacement from the equilibrium position.
Comparing this with the given force F = -25x N, we can equate the two expressions to find k.
-25x = -kx
Dividing both sides by x, we get:
-25 = -k
k = 25 N/m
The angular frequency, ω, is given by ω = √(k/m).
Substituting the values, we have:
ω = √(25/0.25) = √100 = 10 rad/s
Step 2: Finding the amplitude, A
The maximum speed of the particle, vmax, is related to the amplitude, A, and angular frequency, ω, by the equation:
vmax = Aω
Substituting the given values, we have:
4 = A * 10
Dividing both sides by 10, we get:
A = 4/10 = 0.4 m
Converting the amplitude from meters to centimeters:
Amplitude in cm = 0.4 * 100 = 40 cm
Therefore, the amplitude of the motion is 40 cm.
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A particle of mass 250 g executes a simple harmonic motion under a periodic force F = (-25x)N.The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.Correct answer is '40'. Can you explain this answer?
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