Question

The time period T of a spring mass system depends upon mass m and spring constant k and length of spring l (k=force/length).Find relation among T,m,l,k

Answer 1

Given :
T= Time period 
m= mass
l= length 
K = spring constant ( inertia factor ) 

As we know :

F= -ky 

 F=-m$ω^{2}$y

$ω ^{2}$=$\frac{k}{m}$

ω =√k/m 

T= 2π/ω

T= 2π√m/k 

T= 2π√$\frac{Inertia factor }{spring factor }$

Inertia factor = means of inertia of bob about the point of suspension
= m$l^{2}$

Spring factor = mgl 

T= 2π√m$\frac{ l^{2} }{mgl}$ 

T= 2π√l/ g

It means

T α √ m 

T α  1/ √k

T α √l

May it helps you ^ _ ^

Answer 2

Concept:

We will find the relation between T, m, l, k by using dimensional analysis and principle of Homogeneity.

Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same.

Given:

Time period depend upon mass(m) and spring constant(k), length of spring (l), K=force/length.

Find:

We have to find the relation among T, m, k, l.

Solution:

T = a[M]^p

T = a[M T^-2]^q       ∵[k] = [M T^-2}

T = a[L]^r

By using principle of homogeneity,

T = a [M]^p [M T^-2]^q [L}^r --------- 1

[M^0 L^0 T^1 ] = a [M]^p+ q  [L]^r  [T]^-2q

On equating both sides we get,

p+ q = 0 ,   -2q = 1 , r =0

p = -q ,         q = -1/2

p = 1/2 , q= -1/2 , r= 0

on putting the value of p, q, r in equation 1 , we get

T = a m^1/2 k^-1/2

   = a $\sqrt{\frac{m}{k} }$

  From the experiment the value a will be 2π.

  ∴ T = 2π$\sqrt{\frac{m}{k} }$

Hence the relation comes like this.

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