Springs in series and parallel – problems and solutions

Lò xo mắc nối tiếp và song song – các vấn đề và giải pháp

1. Một vật nặng 160 gam được gắn vào một đầu của lò xo và độ thay đổi chiều dài của lò xo là 4 cm. Hỏi độ thay đổi chiều dài của ba đầu lò xo là bao nhiêu? lò xo mắc nối tiếp và song songnhư hình dưới đây?

Đã biết:

Sự thay đổi chiều dài của lò xo (Δx) = 4 cm = 0.04 m Lò xo mắc nối tiếp và song song – các vấn đề và lời giải 1

Thánh Lễ (m) = 160 gam = 0.16 kg

Gia tốc do trọng lực (g) = 10 m/s²

Trọng lượng máy (w) = mg = (0.16)(10) = 1.6 Newton

Muốn : The change in length of three spring (Δx)

Giải pháp:

Phương trình của Luật Hooke :

k = w / Δx = 1.6 / 0.04 = 40 N/m

The three springs have the same constant, k = 40 N/m.

Determine the equivalent constant :

Mùa xuân 2 (k)₂) và mùa xuân 3 (k)₃) tare connected in parallel. The equivalent constant :

k₂₃ = k₂ +k₃ = 40 + 40 = 80 N/m

Mùa xuân 1 (k)₁) và mùa xuân 23 (k)₂₃) are connected in series. The equivalent constant :

1/k = 1/k₁ + 1/k₂₃ = 1/40 + 1/80 = 2/80 + 1/80 = 3/80

k = 80/3

Determine the change in length of three springs :

Δx = w / k = 1.6 : 80/3 = (1.6)(3/80) = 4.8 / 80 = 0.06 m = 6 cm

2. Three springs with the same constant connected in series and parallel, and a 2-kg object attached at one end of a spring, as shown in figure below. Spring constant is k₁= k₂ = k₃= 300 N/m. What is the change in length of the three springs. Acceleration due to gravity is g = 10 m.s^-2.

Đã biết:

Spring constant k₁= k₂ = k₃= 300 Nm^-1

Gia tốc trọng trường (g) = 10 m/s²^-2

Khối lượng của vật (m) = 2 kg

Object’s weight (w) = m g = (2)(10) = 20 Newton

Xem thêm Chu trình Carnot – các vấn đề và giải pháp

Muốn : The change in length of the three springs (Δx)

Giải pháp:

Determine the equivalent constant :

Mùa xuân 1 (k)₁) và mùa xuân 2 (k)₂) are connected in parallel. The equivalent constant :

k₁₂ = k₁ +k₂ = 300 + 300 = 600 N/m

Mùa xuân 3 (k)₃) và mùa xuân 12 (k)₁₂) are connected in series. The equivalent constant :

1/k = 1/k₃ + 1/k₁₂ = 1/300 + 1/600 = 2/600 + 1/600 = 3/600

k = 600/3 = 200 N/m

Determine the change in length of the three springs :

Δx = w / k = 20/200 = 2/20 = 1/10 = 0.1 m

3. Three springs are connected in series and parallel, as shown in figure below. If spring constant k = 50 Nm^-1 and a mass of 400 gram attached at one end of a spring. What is the change in length of the three springs.

Đã biết:

Spring constant 1 (k₁) = k = 50 Nm^-1

Spring constant 2 (k₂) = k = 50 Nm^-1

Spring constant 3 (k₃) = 2k = 2 (50 Nm^-1) = 100 Nm^-1

Object’s mass (m) = 400 gram = 0.4 kg

Gia tốc trọng trường (g) = 10 m/s²

Object’s weight (w) = m g = (0.4)(10) = 4 Newton

Muốn : The change in length (Δx)

Giải pháp:

Determine the equivalent constant :

Mùa xuân 1 (k)₁) và mùa xuân 2 (k)₂) are connected in parallel. The equivalent constant :

k₁₂ = k₁ +k₂ = 50 + 50 = 100 N/m

Mùa xuân 3 (k)₃) và mùa xuân 12 (k)₁₂) are connected in series. The equivalent constant :

1/k = 1/k₃ + 1/k₁₂ = 1/100 + 1/100 = 2/100

k = 100/2 = 50 N/m

Determine the change in length of the three springs :

Δx = w / k = 4 / 50 = = 0.08 m = 8 cm

How does combining springs in series affect the overall spring constant?
- Câu trả lời: When springs are combined in series, the overall spring constant is reduced. The reciprocal of the equivalent spring constant is the sum of the reciprocals of the individual spring constants: 1/k $= 1 / k ^{1} + 1 / k ^{2}$ .
How does the overall spring constant change when springs are combined in parallel?
- Câu trả lời: Combining springs in parallel results in an overall spring constant that is the sum of the individual spring constants: k $= k_{1} + k_{2}$ .

If two identical springs are arranged in series, how does the combined spring constant compare to the spring constant of an individual spring?
- Câu trả lời: The combined spring constant will be half of the spring constant of one of the individual springs.

What happens to the extension or compression of springs in series when a force is applied?
- Câu trả lời: For springs in series, the same force causes each spring to extend or compress, but the total extension (or compression) is the sum of the extensions (or compressions) of the individual springs.

If springs in parallel are subjected to a force, how is that force distributed?
- Câu trả lời: For springs in parallel, the force is distributed among the springs based on their spring constants. Springs with a higher spring constant will bear a greater portion of the force than those with a lower spring constant.
Why can springs in series be thought of as a single spring with a longer length?
- Câu trả lời: Springs in series have a combined effect equivalent to stretching a single longer spring. The extensions or compressions of the individual springs add up, just as they would in a longer singular spring.

How does the potential energy stored in springs in series compare to that in springs in parallel for the same applied force?
- Câu trả lời: Springs in series store more potential energy than springs in parallel for the same applied force because they undergo a greater combined extension or compression.
If one of the springs in a parallel configuration breaks or becomes ineffective, what happens to the overall behavior of the system?
- Câu trả lời: If one spring in a parallel configuration breaks, the remaining springs will still function. However, the overall spring constant of the system will decrease, and the system won’t be able to exert as much restoring force as before.

Why are springs in series more susceptible to larger deformations than those in parallel for the same applied force?
- Câu trả lời: For springs in series, the same force acts on each spring, causing each one to extend or compress. The total deformation is the sum of the individual deformations. In parallel, the force is distributed among the springs, so each one experiences a reduced effective force, leading to smaller individual deformations.
In practical applications, why might engineers choose to use springs in parallel rather than in series?

Câu trả lời: Engineers might choose springs in parallel to achieve a higher overall spring constant, resulting in stiffer behavior. This setup can also provide redundancy; if one spring fails, the system continues to function, albeit with a reduced overall spring constant.