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PHYSICS 9B MIDTERM 1 October 23, 2008
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Problem 1. (2 points) What happens with the scale reading after you immerse an object into the water?
A. Scale reading will increase by the value of the buoyant force. B. Scale reading will decrease by the value of the buoyant force. C. Scale reading will remain constant. D. Scale reading will increase by the weight of the object. E. Scale reading will decrease by the weight of the object.
Solution: Scale reading will increase by the value of buoyant force. When we immerse the block into the water, there is a buoyant force from the water to the block. Accordingly, when two bodies interact, there is an equal downward force from the block acting on the water.
Problem 2. (2 points) Which of the following represents a wave function ( x is coordinate, t is time, v is velocity)
A. y(x,t) =sin( vx-t ) B. y(x,t) =log( xt-v ) C. y(x,t) =exp( vt+x ) D. y(x,t)= tan (t+v/x) E. y(x,t)= cos (x-t/v)
Solution: Any function f(x+vt) or f(x-vt) is the solution of the wave equation. Answer C.
0.00 kg
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Problem 3. (2 points) A sound source and a listener are both at rest but a strong wind is blowing from the source towards the listener. Compared to the source frequency the listener will hear the sound of
A. Higher frequency B. The same frequency C. Lower frequency
Solution: Let’s consider a coordinate system associated with the wind. Then, both the source and the listener move with the same velocity in one direction. fL=(v+vL)/λbehind =(v+vL)/ (v+v (^) S )fS=fS. The answer is the same frequency and there is no Doppler effect.
Problem 4. (2 points) An ice cube floats in a glass of water. When the ice melts will the glass water level
A. Rise B. Fall C. Remain unchanged
Solution: The water level will remain unchanged. Let’s denote volume of water without the ice cube as Vw. When the ice cube is immersed, it is partially under the water, because its weight should be compensated by the buoyant force. Let’s denote the full volume of ice as Vice and its volume under the water as V’. B= ρwV’g = ρiceViceg which produces V’=Viceρice/ρw. as a part of the ice volume under the water. Therefore the water level (when the ice floats) is now Vw+V’. When the ice melts completely it adds the volume Vadd =mice/ρw=Viceρice/ρw and the volume level is determined by Vw +Vadd. As we see Vadd =V’ therefore the water level should remain unchanged.
Problem 5. (3 points) The hypodermic syringe contains a medicine with the same density of water. The barrel of the syringe has a cross-sectional area of 4.65349 × 10−^5 m^2. The cross-sectional area of the needle is 1.65082 × 10 −^8 m^2. In the absence of a force on the plunger, the pressure everywhere is 1.0 atm. A force of magnitude 1.42268 N is exerted on the plunger, making medicine squirt from the needle. Find the medicine’s flow speed through the needle. Assume that the pressure in the needle remains at atmospheric pressure, that the syringe is horizontal, that the speed v 1 of the fluid in the syringe is much smaller than the speed v 2 of the fluid in the needle and that the speed of the emerging fluid is the same as the speed of the fluid in the needle.
Problem 6. (3 points) A loud factory machine produces sound having a displacement amplitude of 1 μm but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10 Pa. The bulk modulus of the air is 1.42 x 10 5 Pa, the speed of sound in air is 344 m/s.
a) What is the highest frequency to which this machine can be adjusted without exceeding prescribed limit? b) What is the average intensity produced by the sound waves of this machine at this frequency? c) What is the sound intensity level in decibels?
Solution:
a) since p (^) max =BkA=B2πf A/v
f=p (^) max v/(2πAB)=10344/(1.42*10^5 * 6.28 * 10-6^ ) = 3.86 * 10^3 Hz
b) I=0.5 vp (^) max^2 /B=0.5344100/(1.42*10^5 )=0.121 W/m^2
c) β=10log(I/I 0 )= 10log(0.121/10-12^ )=110.8 dB
Problem 7 (3 points) A travelling wave on a string is described by the wave function y(x,t)=A cos( kx- ω t ) where
A=0.5 m, k=0.1 m-1^ and ω=500 rad/s.
a) Find wavelength, period of oscillations, and phase speed of the wave b) Find velocity of the particles of the string as a function of x and t. What is the maximum velocity? c) Find acceleration of the particles of the string as a function of x and t. What is the maximum acceleration?
a) λ =2 π /k =6.28/0.1=62.8 m
T=2 π / ω = 6.28/500=0.01256 s
v= λ /T =5000 m/s
b) v(x,t)= A ωsin( kx- ω t ), v max =Aω=0.5*500=250 m/s
c) a(x,t)=-A ω^2 cos( kx- ω t ), a max =Aω^2 =0.5500500= 125000 m/s
Problem 8. (3 points) You are approaching a police car at 68.1 mph and the police car is approaching you at 94. mph. The wind is blowing from behind you at 53.1 mph toward the police car. Assume that the speed of sound is 700 mph.
a. If the police car has a 1224 Hz siren, what frequency do you hear? b. If the police car passed you, what frequency do you hear now?
When the police car passed you,
f=1224 x [700 - (68.1-53.1)]/[700 + (94.8 + 53.1)] = 989 Hz
vw
vo vs ((((( λfront
vw
vs vo ))))) λbehind
