Suppose an object of mass m is attached to the end of a spring hanging from the

Question

Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. We say that the mass is at its equilibrium position y-0 when the spring hangs at rest. Suppose you push the mass to a position y0 units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is given by the equation (undefined) where k is a constant measuring the stiffness of the spring (the larger the value of k, the stiffer the spring) and y is positive in the upward direction.

Questions:

-How would the velocity be affected if the experiment were repeated with 4 times the mass on the end of the spring?

-How would the velocity be affected if the experiment were repeated with a spring having 4 times the stiffness (k is increased by a factor of 4?)

-Find the second derivative d^2y/dt^2 and verify that d^2y/dt^2=-(k/m)y

-After finding the period  2pi (sqrt m/k) assume k is a constant and calculate dT/dm

Explanation / Answer

in simple harmonic oscillations,

displacement(position) of the particle

y=y0*sin(wt)

where

y0 is the amplitude(maximum displacement)

w^2=k/m

"k" is the spring constant

and "m" is the mass

now velocity,

v=dy/dt=w*y0*cos(wt)

or

v^=w^2*(y0^2-y^2)...........(A)

1)

let m1,v1 are initial mass and velocity

and m2,v2 are the final mass and velocity

then m2=4m1(given data)

and from eqation (A),

we can write

v1^2/v2^2=w1^2/w2^2

v1^2/v2^2=(k/m1)/(k/m2)

v1/v2=sqrt(m2/m1)

v1/v2=sqrt(4m1/m1)

v1/v2=2

v2=v1/2

hence final velocity decreases

and becomes the half of the initial velocity

2)now let k1 is initial spring constant

and k2 is final spring constant

then k2=4k1

now v1^2/v2^2=w1^2/w2^

v1^2/v2^2=(k1/m)/(k2/m2)

v1^2/v2^2=(k1/4k1)

v1/v2=sqrt(1/4)

v1/v2=1/2

therfore v2=2*v1

hence final velocity increases

and becomes the twice of the initial velocity

3)

let y=y0*sin(wt)

dy/dt=w*y0*cos(wt)

and d^2y/dt^2=-(w^2*y0*sin(wt))

but w^2=(k/m)

d^2y/dt^2=-(k/m)*y0*sin(wt)

then d^2y/dt^2=-(k/m)*y

henc above eqation is verified

4)

time period T=2pi sqrt( m/k)

if we derivative with respect "m"

dT/dm= pi/sqrt(m*K)   

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