Answers  Simple Harmonic Motion


1.

(a)

The maximum displacement from the
equilibrium position x_{m} = 10.0 cm.


(b)

The time for one complete oscillation
T = π/2 s. Notice
the maximum positive displacement
x = +10.0 cm occurs at t = 0 and the next time at t
= π/2 s.
It occurs again at
t = π s.



2.

(a)

v(t) =
(2 π/T)(x_{m} sin
2 πt/T). The maximum value
of the sine is 1. The maximum absolute value of v =
2 πx_{m}/T. The ±
signs account only for the direction of the velocity.


(b)

From Fig.1b, v_{max} =
40.0 cm/s = 2 πx_{m}/T
=
2 π(10 cm)/( πs/2).



3.

(a)

(i) a(t) =
(2 π/T)^{2}(x_{m}
cos 2 πt/T). Maximum
value of cosine is 1.
a_{max} = (2 π/T)^{2}x_{m}.
The ± signs account for the direction of acceleration.
(ii) a(t) =
(2 π/T)^{2}
(x_{m} cos 2 πt/T)
= (2 π/T)^{2}
x(t), since x(t) = x_{m} cos 2 πt/T.
a(t)/x(t) = (2 π/T)^{2}.


(b)

From Fig.1c, a_{max} =
160 cm/s^{2} = (2 π/T)^{2}x_{m}
= (2 π/( π
s/2)^{2}10cm.



4.

(a) For a mass attached to a spring,
F = kx or F/x = k, where k is
a constant. The applied force F is directly proportional
to the displacement x and the minus sign says it is
in the opposite direction to x.
(b) When the spring is extended and released, F_{net }=
ma or kx = ma or a/x = k/m.


5.

From 3a(ii), a(t)/x(t) = (2 π/T)^{2}.
From 4b, a/x = k/m.
By comparison, k/m = (2 π/T)^{2}
or T= 2 π(m/k)^{1/2}.


6.

(a) Given x(t) = x_{m} cos (2 πt/T +φ), where x_{m} is the maximum displacement
from the equilibrium position. Maximum value of cos (2 πt/T +φ) is 1. The equation accurately describes
the definition of x_{m}.
(b) x(t + T) =x_{m} cos [2 π(t
+ T)/T +φ] = x_{m}
cos T/T +φ] =
A cos [2 πt/T
+ 2π + φ]
= x_{m} cos (2 πt/T +φ) = x(t). T is accurately described by
the equation of motion for simple harmonic motion x_{m}
cos (2 πt/T +φ). It allows for the value of x at t to equal
x at t + T or t + nT,
where n = 1, 2, 3 . . . .
(c) For x(t) = x_{m} cos (2 πt/T
 π/2), x(0) = x_{o}
= x_{m} cos ( π/2)
= 0.
v(t) = (2 π/T)x_{m}
sin (2 πt/T  π/2).
v(0) = v_{o }= (2 π/T)x_{m}
sin (  π/2) = (2 π/T)x_{m}
= +v_{max}
At t = 0, the object is at the equilibrium position and travelling
with the maximum velocity in the +Xdirection.


7.

In general,
x(t) = x_{m}
cos (2 π t/T + φ),
and
v(t) = dx/dt
= x_{m}(2 π/T) sin
(2 π t/T + φ)


(a)

In Fig. for #7a, x(t) is plotted
for φ = 0, and

(d)

in Fig. for #7d, v(t) is plotted
for φ = 0.
For φ = 0 and t = 0 , the
initial conditions are: the initial displacement = x_{o}
= x_{m} and the initial velocity = v_{o}
= 0. Immediately after t = 0, the object moves to the
left (with a negative velocity).

(b)

For φ
=  π/2.
x(t) = x_{m} cos (2 πt/T
 π/2).
By trigonometric identity, cos (C  D) =
cos C cos D + sin C sin D, with C = 2πτ/T
and D = π/2, x(t)
= x_{m}[cos 2πt/T
cos π/2 +
sin 2 πt/T
sin π/2]x(t) = x_{m}[(cos
2πt/T)(0)
+ (sin 2πt/T)(1)] = x_{m}
sin 2 πt/T. Notice the first
maximum in Fig. b lags that in Fig. a by π/2
radians.

(e)

v(t) = 2 π
x_{m}/T sin (2 π
t/T  π/2).
By trigonometric identity, sin (C  D) =
sin C cos D  cos C sin D, v(t) =
2 π x_{m}/T[(sin
2 πt/T)(0)  (cos2 πt/T)(1)]
v(t)
= (2 π x_{m}/T)
cos 2 πt/T. For φ
=  π /2 and
t = 0, x_{o} = 0 and v_{o} = +2 π
x_{m}/T.

(c)

For φ
=  π, x(t) = x_{m}
) =
x_{m}[cos 2 πt/T
cos π + sin 2 πt/T
sin π] =
x_{m}[(cos 2 πt/T)(1)
+ (sin 2 πt/T)(0)]
x(t) =
x_{m} cos 2 πt/T.
Notice that Fig. c lags Fig. a by π radians.

(f)

v(t) = x_{m}(2 π/T)
sin (2 πt/T  π)
=
(2 πx_{m}/T)[sin
2 πt/Tcos π
cos 2 πt/T sin π]
=(2 πx_{m}/T)[(sin
2 πt/T)(1) (cos
2 πt/T)(0)].
v(t) = (2 πx_{m}/T)
sin 2 πt/T. For φ
= π and t = 0, x_{o}
= x_{m} and v_{o} = 0. Immediately
after t = 0, the object moves to the right with a positive
velocity.

(g)

The usefulness of φ
is to state the initial conditions. Note:φ = + π and  π
give the same result.



8.

When the object comes to rest, the net
force acting on it is zero. The magnitude of the force due
to the spring equals the gravitational force,
kx = mg, or k = mg/x = 0.50 kg(9.8 m/s^{2})/0.10 m
= 49 N/m.


9.

In Fig. for #9a, the spring is not stretched.
In Fig. for #9b, the mass is attached and the spring is stretched
a distance of x_{o}.
(a) The mass comes to rest and F_{net }= ma = m(0).
Taking down to be positive,
kx_{o} + mg = 0 (Equation
1)
(b) In Fig. for #9c, the spring has been displaced an additional
distance x.
Now F_{net} = ma, where a ≠ 0 once the spring
is released. Taking the direction of x, which is down, as
positive,
kx  kx_{o} + mg =
ma (Equation
2)
From Eq. 1, we see that  kx_{o} + mg = 0. Eq. 2 becomes
kx = ma or a/x = k/m. The ratio of a to x is the same whether
the spring is mounted horizontally or vertically.
(c) As before, a/x = (2 π/T)^{2
}= (2 πf)^{2}
= k/m and f= (1/2 π)(k/m)^{1/2}.


10.

(a) The forces acting on the pendulum bob are itsweight mg
and the tension T in the string.
(b) The only force tangent to the path is a restoring force
mg sin Θ. From the triangle
with distances, we find that sin Θ= x/L and mg sin Θ
= (mg/L)x.
For small displacements, x ≈s and we can think
of the displacement and the restoring force acting horizontally.
F_{net}

=

ma

(mg/L)x

=

ma

(c) Since m, g, and L are constants,
the restoring force, (mg/L)x, is directly proportional to
the displacement and in the opposite direction. The pendulum
is an example of simple harmonic motion. a/x = (g/L) = (2 π/T)^{2
}= (2 πν
)^{2}.
ν = (1/2 π)(g/L)^{1/2}.


11.




(a)

x(t) = x_{m} cos 2 πt/T

(b)

v(t) = dx/dt = (2 π/T)x_{m}
sin 2 πtT

(c)

a(t) = dv/dt =  (4 π^{2}/T^{2})x_{m}
cos 2 πt/T

(d)

U(t) = 1/2 kx^{2 }= 1/2
kx_{m}^{2 } cos^{2}2 πt/T

(e)

K(t) =1/2 mv^{2} = 1/2
(m4 π^{2}/T^{2})x_{m}^{2
}sin^{2 }2 πt/T
= 1/2 (k)x_{m}^{2 }sin^{2 }2 πt/T,
since k/m = 4 π^{2}/T^{2}.

(f)

E(t) = U(t) + K(t) = 1/2 kx_{m}^{2}(cos^{2}
2 πt/T + sin^{2}
2 πt/T) = 1/2 kx_{m}^{2}
= a constant. U is a maximum twice each period
when x = x_{m} and v = 0.
K is a maximum each period when x = 0 and v =2 πx_{m}/T.
U + K = E = constant.



12.

E = 1/2 kx_{m}^{2} =
1/2(10 N/m)(0.050 m)^{2} = 0.0125 J.
U(x) = 1/2 kx^{2} = 1/2(10 N/m)(0.010 m)^{2}
= 0.0005 J.
K = 1/2 mv^{2}.
E = U + K
0.0125 J = 0.0005 J + 1/2 (0.60 kg)v^{2}.
v^{2} = 2/0.60 kg[0.0125  0.0005]J
0.04 Nm/kg = 0.04 kgm^{2}/s^{2}kg_{
}= 0.04 m^{2}/s^{2}.
v = 0.2 m/s


13.

Given x(t) = 0.01 m cos (0.02 π
s^{1} t  π/2)
compare
with
x(t)
= x_{m} cos (2 πt/T
+ φ) and
find
(a) the amplitude x_{m} = 0.01 m (b) 2 π/T
= 0.02 π s^{1};
(b) period T = 100 s,
(c) the frequency ν =
1/T = 0.01 s^{1} and
(d) the initial phase φ
=  π/2.


14.

From Fig. 3 above, we see that
(a) the cosine curve repeats itself every 4.0 s so the period
T = 4.0 s, and
(b) the amplitude of the motion x_{m} =10.0 cm.
(c) If we write the equation of motion as a function of the
cosine, we let φ= 0.
x(t) = x_{m} cos 2 πt/T
= 10.0 cm cos 2 πt/4.0
s = 10.0 cm cos πt s^{1}.
(d) v(t) = dx/dt = 10.0 π/2
cm/s sin πt s^{1}/2.
v_{max} = 5.0 π
cm/s.
(e) a(t) = dv/dt = 10.0 ( π/2)^{2}cm/s^{2
}cos πt s^{1}/2.
_ a_{max}_
= 2.5 π^{2} cm/s.


15.

For a total swing back and forth of
4.0 cm, the amplitude x_{m} is 2.0 cm.
For x(t) = x_{m} cos (2 πt/T
+ φ), v(t) = dx/dt
= x_{m}2 π/T sin
(2πt/T + φ).
The maximum velocity 10 cm/s occurs at the center of the swing
when x = 0. When cos (2 πt/T
+ φ) = 0, sin (2 πt/T
+ φ) = 1 and v_{max
}= 2πx_{m}/T.
T = 2 πx_{m}/v_{max}=
2 π(2.0 cm)/10 cm/s = 0.4 π
s.


16.

x(t) = 4.0 cm cos ( πt
s^{1 } π/6).
2.0 cm = 4.0 cm cos ( πt
s^{1 } π/6).
cos ( πt s^{1 }
π/6) = 0.5 and ( πt
s^{1 } π/6)
= π/3 (or 60^{0}).
v(t) = dx/dt = 4.0 π cm/s
sin ( πt s^{1 } π/6).
When ( πt s^{1 }
π/6) = π/3,
sin π/3 = 0.866,
and
v = 4.0 π(0.866) cm/s = 10.9 cm/s.


17.

(a)

τ
= r x F



τ
= rF sin _r, F

About the pivot point, the torque for
the rod of mass m_{1} is (L/2)m_{1}g sin
Θ and for the point mass
it is Lm_{2}g sinΘ.
The negative signs occur because they are restoring
torques. When the pendulum is swinging counterclockwise, the
torque tends to make it swing clockwise. The moment of inertia
of the rod about an end is 1/3 m_{1}L^{2}.
The moment of inertia of a point particle of mass m_{2}
a distance L from the axis is m_{2}L^{2}.
(b)

t
= Ia


Lg(m_{1}/2
+ m_{2})sin Θ
= (m_{1}/3 + m_{2})L^{2}
d^{2}Θ/dt^{2}


For small Θ,
sin Θ
is approximately equal to Θ
so that


Lg(m_{1}/2
+ m_{2}) Θ
= (m_{1}/3 + m_{2})L^{2}
d^{2}Θ/dt^{2}


or
d^{2}Θ/dt^{2
}+ [g(m_{1}/2 + m_{2})/(m_{1}/3
+ m_{2})L]Θ
= 0

(c) Comparing with d^{2}x/dt^{2 }+ [k/m]x
= 0 for which T = 2 π (m/k)^{1/2},
we find for this pendulum, T = 2 π
[2(m_{1} + 3m_{2})L/(m_{1} + 2m_{2})3g]^{1/2}


18.

F_{net }= ma = m d^{2}x/dt^{2}


(a) For Fig. 5a,
k_{1}x  k_{2}x = (k_{1}
+ k_{2})x = m d^{2}x/dt^{2}, or
d^{2}x/dt^{2 }+ [(k_{1}
+ k_{2})/m]x = 0
Compare with,
d^{2}x/dt^{2 }+ [k/m]x = 0, when
ν = (1/2 π)(k/m)^{1/2}
and
find for this case,
ν = (1/2 π)[(k_{1}
+ k_{2})/m]^{1/2}
The"effective" spring constant for springs in parallel
is
k_{eff}= k_{1} + k_{2 }.
. + k_{n}
(b) For Fig. 5b, the spring with constant k_{2} is in
contact with mass m that has a displacement x = x_{1}
+ x_{2}, where x_{1} is the extension of the
spring with constant k_{1} and x_{2} the extension
of the spring with constant k_{2}. The force on the
object is,
k_{2}x_{2} and k_{2}x_{2}
= m d^{2}x/dt^{2} (Equation1)
Also
x = x_{1} + x_{2 }(Equation
2)
and the magnitude of the force on the second spring due to the
first spring equals the magnitude of force on the first spring
due to the second spring, or
k_{1}x_{1} = k_{2}x_{2}
or x_{1} = k_{2}x_{2}/k_{1 }(Equation
3)
Substituting Eq. 3 into Eq. 2,
x = (k_{2}x_{2}/k_{1})
+ x_{2} = (k_{1} + k_{2})x_{2}/k_{1}
, or
x_{2} = k_{1}x/(k_{1} + k_{2}) (Equation
4)
Substituting Eq. 4 into Eq. 1,
[k_{2}k_{1}/(k_{1} + k_{2})]x
= m d^{2}x/dt^{2 }, or
d^{2}x/dt^{2 }+ [{k_{1}k_{2}/(k_{1}
+ k_{2}}/m)]x = 0
Comparing with,
d^{2}x/dt^{2 }+ [k/m]x = 0
For this case,
ν = (1/2_)[{k_{1}k_{2}/(k_{1}
+ k_{2})}/m]^{1/2}
The"effective" spring constant for series is,
k_{eff} = {k_{1}k_{2}/(k_{1}
+ k_{2}}, or
1/k_{eff} = 1/k_{1} + 1/k_{2} + .
.+ 1/k_{n}.


19.

Imagine the spring cut into thirds with
each part having a spring constant k’.
When the three springs are connected in series, the spring constant
is
k = 10.0 N/m. For series,
1/k = 1/10 N/m = 1/k’ + 1/k’ + 1/k’=
3/k’
k’ = 3k = 30.0 N/m
When two of these springs with k’ are connected in series,
1/k” = 1/k’ + 1/k’ = 2/30 N/m
or the spring constant with 2/3 of the spring left (1/3 cut
off) is
k” = 15 N/m
T = 2 π(m/k”)^{12}
= 2 π(0.30/15)^{1/2 }s
= 0.89 s.


20.




(a) From Fig. 6 above, at x = 0.025 m,
U = 0.050 J.
(b) E = 0.200 J = U + K = 0.050 J + K. K = 0.150
J.
(c) U = 0.050 J = 1/2 kx^{2}= 1/2 k(0.025 m)^{2}, or
k = 0.10 J/6.25 x 10^{4}
m^{2} = 160 N/m.
(d) K = 0.150 J = 1/2 mv^{2} = 1/2(0.30kg)v^{2},
or
v^{2} = 2(0.150)/0.30
m^{2}/s^{2} = 1.0 m^{2}/s^{2}
or v = 1.0 m/s.
(e) E = 0.200 J =1/2 kx_{m}^{2} =1/2 (160
N/m)x_{m}^{2,} or
x_{m} = (25 x 10^{4})^{1/2}m
= 0.05m.
(f) v is a maximum when the potential energy is a minimum.
This occurs for x = 0 and, therefore, U = 0 with K = E = 0.200
J = 1/2 mv_{max}^{2} = 1/2(0.30kg)v_{max}^{2}.
v_{max }= 1.15 m/s.


