Question

Two springs having stiffness 20N/mm and 25 N/mm are attached in series and a mass is suspended at the end of it. The equivalent spring stiffness of the two springs is nearly ​

Answer 1

Answer:

ANSWER

Series combination :

Equivalent spring constant:

⇒k

s

=

k

1

+k

2

k

1

×k

2

=

k+k

k×k

=

2

k

Time period, T

s

=2π

k

s

m

=2π

k

2m

............(1)

Parallel combination :

Equivalent spring constant:

⇒k

p

=k

1

+k

2

=k+k=2k

Time period, T

p

=2π

k

p

m

=2π

2k

m

............(2)

⇒

T

s

T

p

=

2

1/

2

=

2

1

⇒

2

T

p

=

2

1

⇒T

p

=1 s

Answer 2

Concept:

A body's stiffness (k) is a measurement of how resistant an elastic body is to deform. In general, the force necessary to generate unit deformation for a spring is determined by the spring stiffness.

Given:

Two springs in series have the stiffness of 20N/mm and 25 N/mm respectively and a mass is suspended at the end.

Find:

The equivalent spring stiffness.

Solution:

When two or more springs are connected end to end, they are said to be in series, and when they are connected side by side, they are said to be in parallel.

Now, in series connection.

The equivalent spring constant is:

$\frac{1}{k_{eq}} =\frac{1}{k_1} +\frac{1}{k_2}$

$k_{eq}=\frac{k_1k_2}{k_1+k_2}$

Two springs having stiffness 20N/mm and 25 N/mm are attached in series.

Let, k₁ = 20N/mm and k₂ = 25 N/mm

Therefore, the equivalent stiffness is:

$k_{eq}=\frac{(20)(25)}{25+20}$

$k_{eq}=\frac{500}{45}$

$k_{eq}=11.11\text{N/mm}$

The equivalent spring stiffness is 11.11 N/mm.

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