3, tor as a fune (b) Estimate the frequency. v (m/s) +5 0. 0.1 0.2 0.3 -5 Figure P11.11 11.2 EXAMPLES OF SIMPLE HARMONIC MOTION 12. The period of oscillation for a mass-on-a-spring system is 0.22 s. If m = 3.5 kg, what is the spring constant of the spring?

University Physics Volume 1
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Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter15: Oscillations
Section: Chapter Questions
Problem 67AP: A 2.00-kg block lies at rest on a frictionless table. A spring, with a spring constant of 100 N/m is...
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Problem 14. Must use formula from formula sheet. I’m not sure where to start
3,
tor as a fun
(b) Estimate the frequency.
m/s)
0.
0.2
0.3
0.1
-5
Figure P11.11
EXAMPLES OF SIMPLE
HARMONIC MOTION
11.2
12. The period of oscillation for a mass-on-a-spring system is 0.22 s.
If m = 3.5 kg, what is the spring constant of the spring?
13. Consider the mass on a spring in Figure 11.6. If the spring
constant is k
period?
14. * A child plays on a bungee cord and oscillates with a certain
frequency f. An adult with a mass that is five times greater than
that of the child then uses the same bungee cord. What is the ratio
of the frequency with the adult to the frequency with the child?
15. * Suppose Figure P11.3 describes the displacement of a mass-
on-a-spring harmonic oscillator. (a) If the mass is m =
what is the spring constant? (b) Estimate the velocity at t = 4.0 s
and at 5.0 s. (c) Estimate the acceleration at 1.5 s. (d) Estimate
the maximum acceleration.
= 30 N/m and the mass is m =
2.5 kg, what is the
2.0 kg,
Transcribed Image Text:3, tor as a fun (b) Estimate the frequency. m/s) 0. 0.2 0.3 0.1 -5 Figure P11.11 EXAMPLES OF SIMPLE HARMONIC MOTION 11.2 12. The period of oscillation for a mass-on-a-spring system is 0.22 s. If m = 3.5 kg, what is the spring constant of the spring? 13. Consider the mass on a spring in Figure 11.6. If the spring constant is k period? 14. * A child plays on a bungee cord and oscillates with a certain frequency f. An adult with a mass that is five times greater than that of the child then uses the same bungee cord. What is the ratio of the frequency with the adult to the frequency with the child? 15. * Suppose Figure P11.3 describes the displacement of a mass- on-a-spring harmonic oscillator. (a) If the mass is m = what is the spring constant? (b) Estimate the velocity at t = 4.0 s and at 5.0 s. (c) Estimate the acceleration at 1.5 s. (d) Estimate the maximum acceleration. = 30 N/m and the mass is m = 2.5 kg, what is the 2.0 kg,
Subscripts
volume ,no"u" use
MusH start wlone of These eguahions
use "v" for
Physics 104 Equation Sheet
k = 9.0 x 10° Nm²/C²
Ho = 4n x 107 N/A²
C = 3.00 x 10° m/s
e = 1.6 x 1019 C
E0 = 8.85 x 1012 C²/Nm?
"= 6.63 x 10 34 Js = 4.14 x 1015 eVs lo = 1012 w/m?
me = 9.11 x 10³1 kg
m, = 1.67 x 1027 kg
nair = 1.00
nsoap = 1.50
Wsilver = 4.74 eV
nwater = 1.33
Vsound = 340 m/s
1 eV = 1.6 x 10 19 J
F = kqıq2/r²
PE = kqı92/r
V = kQ/r
AV = -EAx
E = kQ/r?
F = qE
C = q/AV
AV = IR
R = pL/A
APEE = -WE = -qEd
APE = qAV
C = E0A/d
PE = ½ QAV
1/Reg = E, 1/R;
P = IAV
Reg = E; R,
| = Aq/At
F = qvB sino
F = ILB sino
B = Hol/2nr
OB = BA cosO
EMF = NA DB/At
B = Ho nl
IV1 = 12 V2
V1/V2 = N1/N2
EMF = NBAW sin(wt)
v = fA
n = c/v
An = Aair/n
f = 1/T
W = 2nf =
%3D
F = -kAx
PE = ½ kAx?
V =
ni sin1 = n2 sino2
fs
fo =
B = 10 log10 (1/lo)
12/l1 = r/r?
1/o + 1/i = 1/f
fo = fs(1 ± °o/p)
d sino = mA
m = -i/o
f = R/2
P = 1/f
d sino = (m + ½ )A
Am T = 2.90 x 10 mK
KEmax = hf – W
E = hf
En = -13.6eV/n?
L = nh/2n
Rn = (0.0529 nm) n²
A = h/mv = h/p
A-X' = (h/mc)(1 – cos©)
AxAp. 2 h/47
ΔΕΔt2 h/4π
Transcribed Image Text:Subscripts volume ,no"u" use MusH start wlone of These eguahions use "v" for Physics 104 Equation Sheet k = 9.0 x 10° Nm²/C² Ho = 4n x 107 N/A² C = 3.00 x 10° m/s e = 1.6 x 1019 C E0 = 8.85 x 1012 C²/Nm? "= 6.63 x 10 34 Js = 4.14 x 1015 eVs lo = 1012 w/m? me = 9.11 x 10³1 kg m, = 1.67 x 1027 kg nair = 1.00 nsoap = 1.50 Wsilver = 4.74 eV nwater = 1.33 Vsound = 340 m/s 1 eV = 1.6 x 10 19 J F = kqıq2/r² PE = kqı92/r V = kQ/r AV = -EAx E = kQ/r? F = qE C = q/AV AV = IR R = pL/A APEE = -WE = -qEd APE = qAV C = E0A/d PE = ½ QAV 1/Reg = E, 1/R; P = IAV Reg = E; R, | = Aq/At F = qvB sino F = ILB sino B = Hol/2nr OB = BA cosO EMF = NA DB/At B = Ho nl IV1 = 12 V2 V1/V2 = N1/N2 EMF = NBAW sin(wt) v = fA n = c/v An = Aair/n f = 1/T W = 2nf = %3D F = -kAx PE = ½ kAx? V = ni sin1 = n2 sino2 fs fo = B = 10 log10 (1/lo) 12/l1 = r/r? 1/o + 1/i = 1/f fo = fs(1 ± °o/p) d sino = mA m = -i/o f = R/2 P = 1/f d sino = (m + ½ )A Am T = 2.90 x 10 mK KEmax = hf – W E = hf En = -13.6eV/n? L = nh/2n Rn = (0.0529 nm) n² A = h/mv = h/p A-X' = (h/mc)(1 – cos©) AxAp. 2 h/47 ΔΕΔt2 h/4π
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