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Grade 12Mechanics

A spring of unstretched length l has a mass m with one end fixed to a rigid support .Assuming spring to be made of a uniform wire,find the kinetic energy possessed by it if its free end is pulled with uniform velocity v .

Vignesh R

11 Years agoGrade 12

5 Answers

sarim khan

11 Years ago

whenever u find such questions replace m of spring by m/3 ..now see how to do it.

k.e=1/2mv^2

the velocity of a point dx at a distance x from the rigid support will be vx/l.

let w=m/l(linear mass density)

so k.e of particle at x distance of dx thickness=1/2* w*dx*(vx/l)^2

int we get

k.e=1/2 *(m/3)*v^2

poopoo

9 Years ago

fucking hate the way indians answer questions.... “replace m of spring by m/3” -like its sorcery

quote: “

whenever u find such questions replace m of spring by m/3 ..now see how to do it.

k.e=1/2mv^2

the velocity of a point dx at a distance x from the rigid support will be vx/l.

let w=m/l(linear mass density)

so k.e of particle at x distance of dx thickness=1/2* w*dx*(vx/l)^2

int we get

k.e=1/2 *(m/3)*v^2”

Moonshine

9 Years ago

We can find the effective mass of the spring by finding its kinetic energy. This requires adding all the length elements' kinetic energy, and requires the following integral:

$T =\int_m\frac{1}{2}u^2\,dm$

Since the spring is uniform, $dm=\left(\frac{dy}{L}\right)m$ , where $L$ is the length of the spring. Hence,

$=\frac{1}{2}\frac{m}{L}\int_0^L u^2\,dy$ $T = \int_0^L\frac{1}{2}u^2\left(\frac{dy}{L}\right)m\!$

The velocity of each mass element of the spring is directly proportional to its length, i.e. $u=\frac{vy}{L}$ , from which it follows:

$T =\frac{1}{2}\frac{m}{L}\int_0^L\left(\frac{vy}{L}\right)^2\,dy$

$=\frac{1}{2}\frac{m}{L^3}v^2\int_0^L y^2\,dy$

$=\frac{1}{2}\frac{m}{L^3}v^2\left[\frac{y^3}{3}\right]_0^L$ hence the answer $=\frac{1}{2}\frac{m}{3}v^2$

Samuel Garry

7 Years ago

We can find the effective mass of the spring by finding its kinetic energy. This requires adding all the length elements' kinetic energy, and requires the following integral:

$T =\int_m\frac{1}{2}u^2\,dm$

Since the spring is uniform, $dm=\left(\frac{dy}{L}\right)m$ , where $L$ is the length of the spring. Hence,

$=\frac{1}{2}\frac{m}{L}\int_0^L u^2\,dy$ $T = \int_0^L\frac{1}{2}u^2\left(\frac{dy}{L}\right)m\!$

The velocity of each mass element of the spring is directly proportional to its length, i.e. $u=\frac{vy}{L}$ , from which it follows:

$T =\frac{1}{2}\frac{m}{L}\int_0^L\left(\frac{vy}{L}\right)^2\,dy$

$=\frac{1}{2}\frac{m}{L^3}v^2\int_0^L y^2\,dy$

$=\frac{1}{2}\frac{m}{L^3}v^2\left[\frac{y^3}{3}\right]_0^L$ hence the answer $=\frac{1}{2}\frac{m}{3}v^2$

Samuel Garry

7 Years ago

This is a very important concept of mass spring otherwise all the things done in our early classes of 11th is of mass Le spring but this is a very good question there is a very yes or no explanation of IT in mini box as this concept is widely used in olympiads level hope you understand about it

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