Suppose an object of mass m is attached to the end of a spring hanging from the
ID: 2845765 • Letter: S
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)