A 50-cm-long spring is suspended from the ceiling. A 330 g mass is connected to the end and held at rest with the spring unstret
1 answer:
Answer:
a)32.34 N/m
b)10cm
c)1.6 Hz
Explanation:
Let 'k' represent spring constant
'm' mass of the object= 330g =>0.33kg
a) in order to find spring constant 'k', we apply Newton's second law to the equilibrium position 10cm below the release point.
ΣF=kx-mg=0
k=mg / x
k= (0.33 x 9.8)/ 0.1
k= 32.34 N/m
b) The amplitude, A, is the distance from the equilibrium (or center) point of motion to either its lowest or highest point (end points). The amplitude, therefore, is half of the total distance covered by the oscillating object.
Therefore, amplitude of the oscillation is 10cm
c)frequency of the oscillation can be determined by,
f= 1/2π
f= 1/2π
f= 1.57
f≈ 1.6 Hz
Therefore, the frequency of the oscillation is 1.6 Hz
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Refer to the diagram shown below.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.
Define the unit vector i to point in the eastern direction, and the unit vector j to point in the northern direction.
The first distance is 184 m west. It is represented by
d₁ = -184 i
The second distance is 220 m at 30° south of east. It is
d₂ = 220(cos 30° i - sin 30° j) = 190.53 i - 110 j
The third distance is 104 m at 80 east of north. It is
d₃ = 104(sin 80° i + cos 80° j) = 102.42 i + 18.06 j
Let the fourth distance be
d₄ = a i + b j
Because the traveler ends back at the original position, the vector sum of the distances is zero. It means that each component of the vector sum is zero.
The x-component yields
-184 + 190.53 + 102.42 + a = 0
a = -108.95
The y-component yields
0 - 110 + 18.06 + b = 0
b = 91.94
The magnitude of the fourth displacement is
√[(-108.95)² + 91.94² ] = 142.56 m
The direction is at an angle θ north of west, given by
θ = tan⁻¹ (91.94/108.95) = 40.2°
Answer:
The fourth displacement has a magnitude of 142.56 m. It is about 40° north of west.