Answers  Simple Harmonic Motion


1.

 The maximum displacement from the equilibrium position
A = 10.0 cm.
 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 = πs.


2.

 v(t) =  (2 π/T)(A
sin 2 πt/T). The
maximum value of the sine is 1. The maximum absolute
value of v = 2 πA/T.
The ± signs account only for the direction
of the velocity.
 From Fig. 1b above, v_{max}
= 40.0 cm/s = 2 πA/T =
2 π(10cm)/ πs/2.


3.

 a(t) = (2 π/T)^{2}(A
cos 2 πt/T).
Maximum value of the cosine is
1. a_{max} = (2 π/T)^{2}A.
The ± signs account only for the direction
of the acceleration.
 a(t) =  (2 π/T)^{2}
(A cos 2 πt/T) =
 (2 π/T)^{2}
x(t)
since x(t) = A cos 2 πt/T.
a(t)/x(t) = (2 π/T)^{2}.
 From Fig. 1c above,
a_{max} = 160.0 cm/s^{2}
= (2 π/T)^{2}A
= (2 π/ πs/2)^{2}
10 cm.


4.

 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.
 When the spring is extended and released, F_{net
}= ma or
 kx = ma or
 a/x = k/m.
 From 3a(ii),  a(t)/x(t) = (2 π/T)^{2
} By comparison, k/m = (2 π/T)^{2}
or
T = 2 π(m/k)^{1/2}.


5.

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


6.

Given that d^{2}x/dt^{2}
+ (k/m)x = 0
(Equation
1")
Show that x(t) = A cos (2πt/T
+ δ) is a solution.
(Equation
1')
v = dx/dt =  (2 π/T)A
sin (2 πt/T + δ).
dv/dt = d^{2}x/dt^{2} =  (2 π/T)^{2}[A
cos (2 π\t/T + δ)]
=  (2 π/T)^{2 }x
(Equation
2)
Substituting Eq. 2 into Eq. 1",  (2 π/T)^{2
}x + (k/m)x = 0.
This equation is true if (2 π/T)^{2
}= (k/m) or
T = 2 π(m/k)^{1/2}.


7.

In general, x(t) = A cos (2 πt/T
+ δ) and
v(t) = dx/dt = A(2 π/T)
sin (2 πt/T + d).
(a) and (d)
In Fig. for #7a above, x(t) is plotted for
δ = 0.
In Fig. for #7d above, v(t) is plotted for
δ = 0.
For δ
= 0 and t = 0 , the initial conditions are:
the initial displacement = x_{o} = A and
the initial velocity = v_{o} = 0.
Immediately after t = 0, the object moves to the
left (with a negative velocity).
(b) For δ =  π/2,
x(t) = A cos (2 πt/T  π/2).
By trigonometric identity,
cos (C  D) = cos C cos D + sin C sin D (with C
= 2 πt/T and D = π/2)
x(t) = A[cos 2 πt/T cos
π/2 + sin 2 πt/T
sin π/2]
x(t) = A[cos 2 πt/T(0)
+ sin 2 πt/T(1)] = A sin
2 πt/T
Notice the first maximum in Fig. b (immediately
above) lags that in
Fig. a (above) by π/2
radians.
(e) v(t) =  2 πA/T
sin (2 πt/T  π/2)
By trigonometric identity,
sin (C  D) = sin C cos D  cos C sin D
v(t) = 2 πA/T[(sin 2 πt/T)(0)
 (cos2 πt/T)(1)]
v(t) = (2 πA/T) cos 2 πt/T
For δ
=  π/2 and
t = 0, x_{o} = 0 and
v_{o} = +2 πA/T
(c) For δ =  π,
x(t) = A cos (2 πt/T
 π)
=
A[cos 2 πt/T cos π
+ sin 2πt/T
sin π]
=
A[(cos 2 πt/T)(1) + (sin
2 πt/T)(0)]
x(t) =  A cos 2 πt/T
Notice that Fig. c (immediately above)
lags Fig. a (above) by π
radians.
(f) v(t) =  A(2 π/T)
sin (2 πt/T  π)
=
 (2 πA/T)[sin 2 πt/T
cos πcos 2 πt/T
sin π]
=( 2 πA/T)[(sin2 πt/T)(1)
 (cos2 πt/T)(0)].
v(t) = (2 πA/T)
sin 2 πt/T
For δ
=  π and
t = 0, x_{o} =  A and
v_{o} = 0.
Immediately after t = 0, the object moves to the
right with a positive velocity.
(g) The function of δ is
to state the initial conditions.
Note: δ = π or
 π gives
the same result.


8.

 x(t) = A cos (ωt +
δ)
x(0) = x_{o} = A cos δ
(Equation
1)
 v(t) = dx/dt =  ωA
sin (ωt + δ)
v(0) = v_{o} = ωA
sin δ
(Equation
2)
Dividing both sides of Eq. 2 by  ω:
 v_{o}/ω =
A sin δ
(Equation
3)
 Dividing Eq. 3 by Eq. 1, tan δ =
 v_{o}/xω_{o}.
 Squaring Eq. 1 and Eq. 3 and adding:
x_{o}^{2} + ( v_{o}/ω)^{2}
= A^{2}(cos^{2} δ
+ A sin^{2} δ)
or
[x_{o}^{2} + ( v_{o}/ω)^{2}]^{1/2
} = A.


9.

In Fig. for #9(a) above, the spring is not stretched. In Fig.
for #9(b) above, the mass is attached and the spring is stretched
a distance x_{o}.
 The mass comes to rest and F_{net }= ma = m(0).
Taking down to be positive,
 kx_{o} + mg = 0 (Equation
1)
 In Fig. for #9(c) above, the spring has been displaced
an additional distance x. Now Fnet = 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.
 As before,  a/x = (2 π/T)^{2}
= (2 πf)^{2}
= k/m and
f = (1/2 π)(k/m)^{1/2}.


10.

 The forces acting on the pendulum bob are its weight mg
and the tension T in the string.
 The only force tangent to the path is a restoring force
 mg sin Θ. From
the triangle with lengths, we find that
sin Θ =
x/L and  mg sin Θ
=  (mg/L)x.
For small displacements, x ≈ s, we can think of the
displacement and the restoring force acting horizontally.
Fnet
= ma
 (mg/L)x = ma
 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 πf)^{2}.
f = (1/2 π)(g/L)^{1/2}.
For another approach, write Θ
= s/L. For small angles Θ
is approximately equal to sin Θ.
The restoring force  mg sin Θ
≈  mgΘ =
m d^{2}s/dt^{2}
= m d^{2}(LΘ)/dt^{2,}
or  (g/L)Θ = d^{2}Θ/dt^{2
}and d^{2}Θ/dt^{2}
+ (g/L)Θ = 0.
This is the same form as d^{2}x/dt^{2} +
(k/m)x = 0, for which T = 2 π
(m/k)^{1/2} and x = A cos (2 π/T).
By comparison with the spring, for the pendulum T = 2 π
(L/g)^{1/2} and f = 1/T = (1/2 π)(g/L)^{1/2}
and Θ = Θ_{max }
cos (2 π/T).


11.

Given x(t) = 0.01 m cos (0.02 π
s^{1} t  π/2)
compare with x(t) = A cos (2πt/T
+ δ) and find:
 the amplitude A = 0.01 m
 2 π/T = 0.02 π
s^{1}; period T = 100 s,
 the frequency f = 1/T = 0.01 s^{1},
and
 the initial phase δ
=  π/2.


12.

From Fig. 3 above, we see that:
 The cosine curve repeats itself every 4.0s so the period
T = 4.0 s.
 The amplitude of the motion A =10.0 cm.
 If we write the equation of motion as a function of the
cosine, we let δ = 0.
x(t) = A cos 2 πt/T =
10.0 cm cos 2 πt/4.0
s = 10.0 cm cos πt s^{1}/2.
 v(t) = dx/dt = 10.0 π/2
cm/s sin πt s^{1}/2
v_{max} = 5.0 π
cm/s
 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^{2}


13.

For the Xcomponent:
(a) x = r cos Θ = r
cos ωt
(b) v_{x} = dx/dt =  ωr
sin ωt
a_{x} = dv_{x}/dt
=  ω^{2 }(r
cos ωt) =  ω^{2}x
(c) a_{x}/x =  ω^{2}
=  4 π^{2}f^{2}
(d) (e) Since the acceleration is directly proportional to the
displacement and in the opposite direction, the motion is simple
harmonic. Remember by Newton's second law of motion the acceleration
is directly proportional to, and in the same direction as, the
force.
(f) a_{x}/x =  ω^{2}
=  4 π^{2}f^{2}^{
}=  k/m or
ω = 2 πf =
(k/m)^{1/2}
For the Ycomponent:
(a) y = r sin Θ = r sin
ωt
(b) v_{y} = dy/dt = ωr
cos ωt
a_{y} = dv_{y}/dt
=  ω^{2}(r
sin ωt) =  ω^{2}y
(c) a_{y}/y =  ω^{2}
=  4 π^{2}f^{2}
(d) (e) The motion is again simple harmonic, and
(f) a_{y}/y =  ω^{2}
=  4 π^{2}f^{2}
=  k/m or
ω = 2 πf
= (k/m)^{1/2}


14.

For a massspring system,
 kx = ma = md^{2}x/dt^{2 } or
d^{2}x/dt^{2 }+ (k/m)x = 0, where the period
T = 2 π (m/k)^{1/2}.
By comparison with
b^{2}d^{2}x/dt^{2 }+ c^{2}x
= 0 or
d^{2}x/dt^{2 }+ (c/b)^{2}x = 0,
we see that in this case T = 2 π
(b/c).


15.

For a total swing back and forth of 4.0
cm, the amplitude A is 2.0 cm.
For x(t) = A cos 2 πt/T,
v(t) = dx/dt =  A2 π/T
sin 2 πt/T.
The maximum velocity of the pendulum occurs at the center of
the swing equal to 10 cm/s.
v_{max} = 2 πA/T.
T = 2 πA/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 (60^{o}).
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.

ω= 2 πf = 2 π
(3/2 s^{1}) = 3 π
s^{1}. x_{o} = 0.25 m and
v_{o} =  1.5 m/s.
We know that 0 < δ
< π/2 because x_{o}
is positive and v_{o} is negative. The initial position
of the object is less than the amplitude, but positive, and
moving toward the equilibrium position with a negative velocity.
As shown in #8 above,
A = [(x_{o})^{2} + ( v_{o}/ω)^{2}]^{1}^{/2
}= [(0.25 m)^{2} + (1.5 m/s/3 π
s^{1})^{2}]^{1/2} = 0.296 m.
tan δ =  v_{o}/ωx_{o}
= 1.5 m/s/(0.25 m)(3 π s^{1})
= 0.637.
tan^{1} 0.637 = 0.18 π
= 32.5^{o}.
x(t) = A cos (2 πt/T + δ)
= 0.296 m cos (3 πt s^{1
}+ 0.18 π).
Plots of position x and the velocity v as a function of t
are shown in 17a and 17b above, respectively. Note x_{o}
= 0.25 m and v_{o} =  1.5 m/s.


18.

F_{net }= ma = m d^{2}x/dt^{2
} 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
f = (1/2 π)(k/m)^{1/2}
and find for this case,
f = (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}.
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}
is 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 }(Equation
1)
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,
f = (1/2 π)[{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
= 2 π (0.02)^{1/2}
s.


20.

 The differential mass dm = m_{s}dy/L
(Fig. 6 above)
 Assuming the velocity v_{y} at y increases
linearly with y from 0 at y = 0
to v at y = L, v_{y} = vy/L.
 dK = 1/2 dm v_{y}^{2} = 1/2(m_{s}dy/L)(vy/L)^{2}.
The effective mass that takes place in the oscillation is
m_{s}/3.
The fraction of the mass is 1/3.


21.



 The slope of force F as a function of the extension x
is
k = (5.0  0)N/(0.50  0)m = 10 N/m.
 For a total mass M and force constant k, the period of
the motion is
T = 2 π(M/k)^{1/2}.
For a spring of effective mass m_{s}’, M
= m + m_{s}’, where m is the variable
mass added to the spring load. For this case
T = 2 π (m
+ m_{s}’)/k]^{1/2} or
T^{2} = (4 π^{2}/k)m
+(4 π^{2}/k)m_{s}’
(Equation
1)
If we compare Eq. 1 to the equation of a line, y = (slope)x
+ (yintercept), we see that T^{2} versus
m should yield a straight line and the intercept on the
T^{2 }axis is (4 π^{2}/k)m_{s}’.
From Fig. 7a above, we find k = 10 N/m. From Fig.
7b above, we find the intercept on the T^{2} axis
= 0.154 s^{2} = (4 π^{2}/k)m_{s}’
= (4 π^{2}/10
N/m)m_{s}’. m_{s}’ = (1.54/4 π^{2})kg
= 0.390 kg and m_{s} = 3 x (0.390 kg) = 0.117 kg
= 117 g.
 When T^{2} = 0 in Eq. 1, 0 = (4 π^{2}/k)m
+ (4 π^{2}/k)m_{s}’
or
m_{s}’ =  m =  (0.0390
kg).
m_{s}’ = 0.0390 kg, agreeing with the value
found in part (b).
 P.S. It helps to have my graphing program that reads
off coordinates.


22.

For simple harmonic motion, a/x =  ω^{2}.
For maximum acceleration, a_{max}
= Aω^{2}.
For motion in the vertical direction, (F_{net})_{y
}= ma_{y} = m(0) or
N  mg = 0 and N = mg.
The frictional force that keeps the block from slipping on the
plate f = µN = µmg = ma = mAω^{2}.
A = µg/ω^{2}
= 0.60(10 m/s^{2})/(1.5 s^{1})^{2 }=
2.7 m.


23.

 For the equilibrium position, dU/dr = 0 = (5a/r^{6})
+ (3b/r^{4}).
Calling r = r_{o} at the equilibrium position,
r_{o} = (5a/3b)^{1/2}.

The reduced mass µ = m/2 and the frequency f = (1/2 π)(k/µ)^{1/2
}=
(1/2 π)[(12b/m)(3b/5a)^{5/2}]^{1/2}.


24.

 U = mgy
 dU/dx = d(mgy)/dx = mg.
d[R  (R^{2}  x^{2})^{1/2}]/dx
= mgx/(R^{2}  x^{2})^{1/2}.
For x < < R, dU/dx = mgx/R. F_{x}
=  dU/dx =  (mg/R)x.
The force F_{x} is directly proportional to the
displacement x and in the opposite direction. The motion
is simple harmonic.
 F_{x}
= ma or
 (mg/R)x = m d^{2}x/dt^{2 }and
d^{2}x/dt^{2} + (g/R)x = 0.
Compare d^{2}x/dt^{2}
+ (k/m)x = 0 (for which T = 2 π(m/k)^{1/2)
}and see for this case T= 2 π(R/g)^{1/2}.


25.

 When x = A, v = 0 and K = 0. In general, E
= U + K = 1/2 kx^{2} + 1/2 mv^{2}. When
x = A, E = 1/2 kA^{2} + 1/2 m(0) = 1/2 kA^{2}.
Since E is a constant, E always equals:
1/2 kA^{2 }= 1/2 kx^{2} + 1/2
mv^{2}
(Equation
1)
 Multiplying Eq. 1 by 2/m gives:
(k/m)A^{2 }= (k/m)x^{2} +^{
}1/2 mv^{2} or^{
} (k/m)(A2  x2) =^{ } v^{2 } and
^{ }v =^{ }dx/dt =^{ }(k/m)^{1/2}(A^{2}
 x^{2})^{1/2}.
 Separating variables dx/(A^{2}  x^{2})^{1/2} =
(k/m)^{1/2} dt.
Let x = A sin Θ, dx
= A cos Θ dΘ.
(A^{2}  x^{2})^{1/2} = A(1  sin^{2}
Θ)^{1/2} = A
cos Θ.
For limits on Θ, x_{o}
= A sin Θ_{o}
and sin Θ_{o}
= x_{o}/A.
The lower limit is sin ^{1 }x_{o}/A and
the upper is sin ^{1 }x/A.
Integrating the above equation gives:
sin ^{1 }x/A + sin ^{1 }x_{o}/A
= (k/m)^{1/2}t or
sin ^{1 }x/A = (k/m)^{1/2}t + sin ^{1
}x_{o}/A.
Letting sin ^{1 }x_{o}/A = δ,
sin ^{1 }x/A = (k/m)^{1/2}t
+ δ or
x(t) = A sin [(k/m)^{1/2}t + δ].


26.

 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 sin Θ.
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}.
 τ = Iα
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 and
d^{2}Θ/dt^{2
}+ [g(m_{1}/2 + m_{2})/(m_{1}/3
+ m_{2})L]Θ
= 0.
Comparing with
d^{2}x/dt^{2 }+ [k/m]x = 0 (for
which Period = 2 π
(m/k)^{1/2)},
we find for this pendulum,
Period = 2 π
[2(m_{1} + 3m_{2})L/(m_{1} + 2m_{2})3g]^{1/2}


27.

For oscillations about pin axis (Fig. 24 above),
 τ =
r x F.
τ = rF sin r,F
= rmg sin Θ ≈
rmg Θ for small
Θ.
The moment of inertia about the pin a distance r from the
center of mass,
I = (1/2 mR^{2} + mr^{2}).
 τ= Iα
= (I)d^{2}Θ/dt^{2
}rmgΘ = (1/2 mR^{2}
+ mr^{2})d^{2}Θ/dt^{2}
 or
d^{2}Θ/dt^{2
}+ [rg/(1/2 R^{2} + r^{2})]Θ
= 0.
Compare with
d^{2}x/dt + [k/m]x = 0 (for which Period
= 2 π[m/k]^{1/2)}
and see for the disk,
Period = 2 π[(R^{2}
+ 2r^{2})/2rg]^{1/2}.
 For a minimum period,
d(Period)/dr = 0 =
2 π(1/2)[(2rg)(4r) 
(R^{2 }+ 2r^{2})2g]/4r^{2}g^{2}][(R^{2}
+ 2r^{2})/2rg]^{1/2 } or
(2rg)(4r) = (R^{2 }+ 2r^{2})2g or
4r^{2 }= R^{2 }+ 2r^{2 } and
2r^{2} = R^{2} or
r = R/(2)^{1/2}.

