Problem 4: A ball of mass m is connected to two rubber bands of length L, cach under tension T, as shown in the Figure. Assume the tension does not change. (a) The ball of mass m is in equilibrium. Draw a free-body diagram on m. (b) Since the ball of mass m is in equilibrium, Eety=0. Use Fnet.y 0 to find the equilibrium position for the ball yo. Hint: sin 0 yo/L, Answer: o = mgL/(27'). T T. L L. т (c) The ball of mass m is now displaced slightly from equilibrium, as shown in the figure below. Draw a free-body diagram on m. (d) Use Newton's 2nd Law: Fnet.y = ma = n to show that the motion of m satisfies the SHM differential equation + wu = 0, where u = y - yo and w = since the downward gravitational force on the ball is larger than the two upward tension forces on the ball. Hint: sin 0 = y/L. Also, from part (b), yo = mgL/(2T). We can rewrite this expression as mg = 2T(yo/L). Use this expression to substitute for mg in Newton's 2nd Law: Fnet.y = -ma = -m 2T For this problem we use a = V mL dt2 -- y T L T. m

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Problem 4: A ball of mass m is connected to two rubber bands of length L, cach under temsion T, as
shown in the Figure. Assume the tension does not change.
(a) The ball of mass m is in equilibrium. Draw a free-body diagram on m.
(b) Since the ball of mass m is in equilibrium, Enet y = 0. Use Fnet.y=0 to find the equilibrium position
for the ball yo. Hint: sin 0=yo/L, Answer: yo = mgL/(2T').
|Yo
L.
T.
L
т
(c) The ball of mass m is now displaced slightly from equilibrium, as shown in the figure below. Draw a
free-body diagram on m.
(d) Use Newton's 2nd Law: Fnet,y = ma =
-m
to show that the motion of m satisfies the SHM
differential equation + w²u = 0, where u = y – Yo and w =
since the downward gravitational force on the ball is larger than the two upward tension forces on the ball.
Hint: sin 0 = y/L. Also, from part (b), yo = mgL/(2T). We can rewrite this expression as mg = 2T(yo/L).
Use this expression to substitute for mg in Newton's 2nd Law: Fnet.y = -ma = -my
| 2T. For this problem we use a = –
V mL
dt2.
g
T
T.
Transcribed Image Text:Problem 4: A ball of mass m is connected to two rubber bands of length L, cach under temsion T, as shown in the Figure. Assume the tension does not change. (a) The ball of mass m is in equilibrium. Draw a free-body diagram on m. (b) Since the ball of mass m is in equilibrium, Enet y = 0. Use Fnet.y=0 to find the equilibrium position for the ball yo. Hint: sin 0=yo/L, Answer: yo = mgL/(2T'). |Yo L. T. L т (c) The ball of mass m is now displaced slightly from equilibrium, as shown in the figure below. Draw a free-body diagram on m. (d) Use Newton's 2nd Law: Fnet,y = ma = -m to show that the motion of m satisfies the SHM differential equation + w²u = 0, where u = y – Yo and w = since the downward gravitational force on the ball is larger than the two upward tension forces on the ball. Hint: sin 0 = y/L. Also, from part (b), yo = mgL/(2T). We can rewrite this expression as mg = 2T(yo/L). Use this expression to substitute for mg in Newton's 2nd Law: Fnet.y = -ma = -my | 2T. For this problem we use a = – V mL dt2. g T T.
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