StudentAidUnit 7: Waves and Vibrations — Numericals
11th Class Physics · Unit 7: Waves and Vibrations
7.1.The speed of a wave on a typical string is 24 ms⁻¹. What driving frequency will it resonate if its length is 6.0 m?
Given
Speed of wave on string
v = 24 ms-1
Length of string
l = 6.0 m
Frequency
f = ?
Formula
Resonance frequency
f = v2l
Working
Substituting
f = 242 × 6.0
f = 2412
Result
Driving frequency
f = 2 Hz
7.2.The lowest resonance frequency for a guitar string of length 0.75 m is 400 Hz. Calculate the speed of a transverse wave on the string.
Given
Length of string
l = 0.75 m
Frequency
f = 400 Hz
Speed of transverse wave
v = ?
Formula
For fundamental frequency
f = v2l
Rearranging
v = 2lf
Working
Substituting
v = 2 × 0.75 × 400
v = 1.5 × 400
Result
Speed of transverse wave
v = 600 ms-1
7.3.A tuning fork A produces 4 beats per second with another tuning fork B. It is found that by loading B with some wax, the beat frequency increases to 6 beats per second. If the frequency of A is 320 Hz, determine the frequency of B when loaded.
7.4.A steel wire hangs vertically from a fixed point, supporting a weight of 80 N at its lower end. The diameter of the wire is 0.50 mm and its length from the fixed point to the weight is 1.5 m. Calculate the fundamental frequency emitted by the wire when it is plucked. Density of steel wire is 7.8×10³ kg m⁻³.
Given
Weight supported
F = T = 80 N
Diameter of wire
d = 0.50 mm = 0.0005 m
Length of wire
l = 1.5 m
Density of steel
rho = 7.8 × 103 kg m-3
Acceleration due to gravity
g = 9.8 ms-2
Formula
Cross-sectional area
A = pileft(d2right)2
Working
Substituting diameter
A = pileft(0.00052right)2
A = 1.96 × 10-7 m2
Formula
Mass per unit length
m' = rho × A
Working
Substituting
m' = 7.8 × 103 × 1.96 × 10-7
m' = 1.53 × 10-3 kg m-1
Formula
Fundamental frequency formula
f = 12lsqrt{Fm'
Working
Substituting values
f = 12 × 1.5 × sqrt{frac{80}{1.53 × 10-3
f = 13 × sqrt{52287.6}
f = 13 × 228.6
Result
Fundamental frequency
f = 76.2 Hz
7.5.Average intensity of sunlight on the surface of the Earth is nearly 500 W m⁻². Determine the amount of energy that falls on a solar panel having an area of 0.50 m² in four hours.
Given
Sunlight intensity
I = 500 W m-2
Solar panel area
A = 0.50 m2
Time
t = 4 hours = 14,400 s
Energy
E = ?
Formula
Power received
P = I × A
Working
Substituting
P = 500 × 0.50
P = 250 W
Formula
Energy
E = P × t
Working
Substituting
E = 250 × 14,400
Result
E = 3.6 × 106 J
7.6.(a) If the intensity of a wave is 16 W m⁻² and the amplitude is 2 m, what is the value of constant k?
(b) If the intensity of a wave is 25 W m⁻² and the constant k is 5 W m⁻¹ what is the amplitude?
Given
(a) Intensity
I = 16 W m-2
(a) Amplitude
A = 2 m
(a) Constant k
k = ?
Formula
(a) Intensity-amplitude relation
k = IA2
Working
(a) Substituting
k = 1622
(a)
k = 164
Result
(a) Constant k
k = 4 W m-4
Given
(b) Intensity
I = 25 W m-2
(b) Constant k
k = 5 W m-1
(b) Amplitude
a = ?
Formula
(b) Amplitude from intensity
A2 = Ik
Working
(b) Substituting
A = sqrt{255
(b)
A = sqrt{5}
Result
(b) Amplitude
A = 2.24 m
7.7.(a) A sound system produces 200 watts of power. If the sound is directed at a crowd with an area of 150 m², what is the intensity of the sound?
(b) A light bulb emits 100 watts of power. If the light is spread out evenly over a sphere with a surface area of 400 m², what is the intensity of the light?
Given
(a) Power
P = 200 W
(a) Area
A = 150 m2
(a) Intensity
I = ?
Formula
(a) Intensity formula
I = PA
Working
(a) Substituting
I = 200150
Result
(a) Intensity of sound
I = 1.33 W m-2
Given
(b) Power
P = 100 W
(b) Area
A = 400 m2
(b) Intensity
I = ?
Formula
(b) Intensity formula
I = PA
Working
(b) Substituting
I = 100400
Result
(b) Intensity of light
I = 0.25 W m-2
7.8.A radio antenna broadcasts 500 watts of power. If the signal is received at a distance of 10 km, what is the intensity of the signal?
Given
Power
P = 500 W
Distance
r = 10 km = 10,000 m
Intensity
I = ?
Formula
Surface area of sphere
A = 4pi r2
Working
Substituting
A = 4pi (10,000)2
A = 4 × 3.14 × 108
A = 1.26 × 109 m2
Formula
Intensity
I = PA
Working
Substituting
I = 5001.26 × 109
Result
Intensity
I = 4 × 10-7 W m-2
7.9.(a) An organ pipe has a length of 1 m. Determine the frequencies of the fundamental and the first two harmonics if the pipe is open at both ends. (Speed of sound in air is 340 m s⁻¹)
(b) An organ pipe has a length of 1 m. Determine the frequencies of the fundamental and the first two harmonics if the pipe is closed at one end. (Speed of sound in air is 340 m s⁻¹)
Given
(a) Pipe length
l = 1 m
(a) Speed of sound
v = 340 m s-1
Formula
(a) Fundamental frequency (open pipe)
f1 = v2l
Working
(a) Substituting
f1 = 3402 × 1
(a)
f1 = 170 Hz
Formula
(a) 1st harmonic (2nd frequency)
f2 = 2f1
(a)
f2 = 2 × 170 = 340 Hz
(a) 2nd harmonic (3rd frequency)
f3 = 3f1
(a)
f3 = 3 × 170 = 510 Hz
Given
(b) Pipe length
l = 1 m
(b) Speed of sound
v = 340 m s-1
Formula
(b) Fundamental frequency (closed pipe)
f1 = v4l
Working
(b) Substituting
f1 = 3404 × 1
(b)
f1 = 85 Hz
Formula
(b) 1st harmonic (3rd frequency, odd multiples)
f3 = 3f1
(b)
f3 = 3 × 85 = 255 Hz
(b) 2nd harmonic (5th frequency, odd multiples)
f5 = 5f1
(b)
f5 = 5 × 85 = 425 Hz
(a) f1 = 170 Hz, f2 = 340 Hz, f3 = 510 Hz quad (b) f1 = 85 Hz, f3 = 255 Hz, f5 = 425 Hz
7.10.A train is approaching a station at 90 km h⁻¹, sounding a whistle of frequency 1000 Hz. What will be the apparent frequency of the whistle as heard by a listener sitting on the platform? What will be the apparent frequency heard by the same listener if the train moves away from the station with the same speed? (Speed of sound is 340 m s⁻¹)
Given
Train speed
vs = 90 km h-1 = 25 m s-1
Whistle frequency
f = 1000 Hz
Speed of sound
v = 340 m s-1
Apparent frequency (approaching)
f' = ?
Apparent frequency (receding)
f'' = ?
Formula
Doppler formula (approaching)
f' = f × vv - vs
Working
Substituting
f' = 1000 × 340340 - 25
f' = 1000 × 340315
f' = 1079.4 Hz
Formula
Doppler formula (receding)
f'' = f × vv + vs
Working
Substituting
f'' = 1000 × 340340 + 25
f'' = 1000 × 340365
f'' = 931.5 Hz
f' = 1079.4 Hz (approaching), f'' = 931.5 Hz (receding)