A Mass
m
is attached to a spring of mass
M
and spring constant
K
.
If the velocity of the mass m varies linearly over the length of the spring when it is given a kick, Let T denote the Time Period of the system.
Find 1 0 0 0 T to the nearest integer.
Details and Assumptions:
∙
m
=
3
K
g
∙
M
=
0
.
5
K
g
∙
K
=
1
0
0
N
/
m
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how does dx move with a velocity vx/L????....
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Since the velocity of the spring varies linearly, the velocity of any elemental particle at a distance x from the fixed end can be given as u = a x + b , where a and b are constants which can be easily determined.
At x = 0 , i.e, at the fixed end u = 0 . Therefore, 0 = 0 + b ⇒ b = 0 .
Similarly, at x = L , i.e, at the point connected to the block u = v . Therefore, v = a L ⇒ a = L v .
Substituting the above values in the expression, we get, u = L v x .
THANK! ! I was really in need of solution to this type of question .....
T = 2PIE [ ROOT OF (m+M/3) / K ]
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The effective mass of the spring used in calculating the time period is M/3. So the time period is 2(pi)((M/3 + m)/K)^(1/2) instead of 2(pi)(m/k)^(1/2) Plugging in the values, you get T=1.11810... =>1000T = 1118.10... To find out how the effective mass is M/3, you can find the kinetic energy if the spring. Consider an element of the spring of length dx at a distance x from the wall, and let the other end of the spring move with a velocity v. Now dx moves with a velocity of vx/L (the total length of the spring being L). The mass dM of the element is Mdx/L. Therefore the kinetic energy of dx, dKE = (1/2)(Mdx/L)(vx/L)^2. Now, integrating from 0 to L, we end up with KE = (1/6)(M)(v)^2 = (1/2)(M/3)(v)^2