双语物理

1. A fisherman’s scale stretches

2.8 cm when a

3.7kg fish hangs from it.(a)What is

the spring constant?(b)What will be the amplitude and frequency of vibration if the fish is pulled down 2.5cm more and released so that it vibrates up and down? Solution: (a) We find the spring constant from the elongation caused by the weight:k=mg/x∆=(3.7 kg)(9.80 m/s2)/(0.028 m)= 1.30⨯103 N/m.

(b) Because the fish will oscillate about the equilibrium position, the amplitude

will be the distance the fish was pulled down from equilibrium: A = 2.5 cm.

The frequency of vibration will be f

=

= = 3.0 Hz

2. A mass m at the end of a spring vibrates with a frequency of 0.88Hz;when an additional 1.25kg mass is added to m,the frequency is 0.48Hz.What is the value of m?

The dependence of the frequency on the mass is f

=

Because the spring constant does not change, we have

f2/f = (m/m2)1/2;

(0.48 Hz)/(0.88 Hz) = [m/(m + 1.25 kg)]1/2, which gives m = 0.53 kg.

3. (a)Determine the length of a simple pendulum whose period is 1.00s,(b)What would be

the period of a 1.00-m-long simple pendulum？

Solution:(a) We find the length from

2 T=

;1.002

s=which gives L = 0.248 m

(b)We find the period from

2 T=

; 2 2.01 T s

==

4. A tuning fork vibrates at a frequency of 264Hz and the tip of each prong moves 1.5mm to either side of center.Calculate(a)the maximum speed and (b)the maximum acceleration of the tip of a prong.

Solution: The angular frequency of the motion is

ω = 2πf = 2π(264 Hz) = 1.66 ⨯ 103 s–1.

(a) The maximum speed is

v max = ωA = (1.66 ⨯ 103 s–1)(1.5 ⨯ 10–3 m) =2.5 m/s.

(b) The maximum acceleration is

a max = ω2A = (1.66 ⨯ 103 s –1)2(1.5 ⨯ 10–3 m) =4.1 ⨯ 103 m/s 2.

5.A mass of 240g oscillates on a horizontal frictionless surface at a frequency of

3.5Hz and with amplitude of

4.5cm.(a)What is the effective spring constant for this motion?(b)How much energy is involved in this motion?

Solution: (a ) We find the spring constant from2πf = (k /m )1/2;

2π(3.5 Hz) = [k /(0.240 kg)]1/2, which gives k = 1.2 ⨯ 102 N/m .

(b ) We find the total energy from the maximum potential energy: E = U max = 212kA = 12

(1.16 ⨯ 102 N/m)(0.045 m)2 = 0.12 J .

6. Two earthquake waves have the same frequency as they travel through the same

portion of the Earth,but one is carrying twice the energy.What is the ratio of the amplitudes of the two waves?

Solution: Because the speed, frequency, and medium are the same for the two waves,

the intensity depends on the amplitude only: 2I A ∝

For the ratio of intensities we have

22211()I A I A =; 221

2()A A =, which gives 21A A = 1.41. 7. Compare(a)the intensities and (b)the amplitudes of an earthquake P wave as it passes two points 10km and 20km from the source.

Solution:We assume that the wave spreads out uniformly in all directions.

(a ) The intensity will decrease as 1/r 2, so the ratio of intensities is

I 2/I 1 = (r 1/r 2)2 = [(10 km)/(20 km)]2 =0.25. (b ) Because the intensity depends on 2A , the amplitude will decrease as 1/r , so the ratio of amplitudes is 2112100.520A r km A r km

=== 8. A violin string vibrates at 294Hz when unfingered.At what frequency will it vibrate if it is fingered one-fourth of the way down from the end?

Solution:From the diagram the initial wavelength is 2L , and the

final wavelength is 3L /2. The tension has not changed,