if a mass of 10 gm is suspended from spring of k=9.8n/m then the extension will be?
Answers
Answer:
Explanation:Conceptual Question 15.2
A pendulum on Planet X, where the value of g is unknown, oscillates with a period T = 2 s. What
is the period of this pendulum if: (a) Its mass is doubled? (b) Its length is doubled? (c) Its
oscillation amplitude is doubled?
15.2. The period of a simple pendulum is T Lg = . 2 / π We are told that 1T =. . 20s
(a) In this case the mass is doubled: 2 1 m m = . 2 However, the mass does not appear in the formula for the period of a
pendulum; that is, the period does not depend on the mass. Therefore the period is still 2.0 s.
(b) In this case the length is doubled: 2 1 L L = . 2
2 2 2 1
1 1 1 1
2 / 2 2
2 /
T LL L g
T LL L g
π
π = = = =
So 2 1 T T = = . =. . 2 2(2 0 s) 2 8 s
(c) The formula for the period of a simple small-angle pendulum does not contain the amplitude; that is, the period is
independent of the amplitude. Changing (in particular, doubling) the amplitude, as long as it is still small, does not affect
the period, so the new period is still 2.0 s.
It is equally important to understand what doesn’t appear in a formula. It is quite startling, really, the first time you realize
it, that the amplitude max ( ) θ doesn’t affect the period. But this is crucial to the idea of simple harmonic motion. Of course,
if the pendulum is swung too far, out of its linear region, then the amplitude would matter. The amplitude does appear in
the formula for a pendulum not restricted to small angles because the small-angle approximation is not valid; but then the
motion is not simple harmonic motion.
Conceptual Question 15.4
The figure shows a position-versus-time graph for
a particle in SHM. (a) What is the phase constant?
Explain. (b) What is the phase of the particle at
each of the three numbered points on the graph?
15.4. (a) A position-vs-time graph plots 0 x t A wt ( ) cos( ) = +. φ The graph of x(t) starts at 1
2
A and is increasing. So at
t = 0, 0 0
1 cos 2 3
A A π = ⇒ =± . φ φ We choose 0 3
π φ = − since the particle is moving to the right, indicating that it is in
the bottom half of the circular-motion diagram.
(b) The phase at each point can be determined in the same manner as for part (a). For points 1 and 3, the amplitude is
again 1
2
A. At point 1, the particle is moving to the right so 1 3
π φ =− . At point 3, the particle is moving left, so 3 .
3
π φ = +
At point 2, the amplitude is A, so 2 2 cos 1 0 φ φ =⇒ =.
Answer:
10 cm
Explanation:
equating the forces acting at equilibrium,
Kx=mg
9.8(x)= 10^(-2)*(9.8)
so x=10 cm