Your cart is empty.
Practice each question, then open the answer or worked solution.
The displacement of an elastic wave is given by the function y = 3 sin ωt + 4 cos ωt where y is in cm and t is in second. Calculate the resultant amplitude.
Given,
y = 3 sin ωt + 4 cos ωt
Let’s consider
3 = a cos φ
4 = a sin φ
Then
y = a cos φ ωt + a sin φ ωt
y = a sin (ωt + φ)
tan φ = 4/3
φ = tan-1 4/3
When the above equation is squared, we het
a2 = 25
a = 5
Therefore, the new amplitude is 5 cm.
A sitar wire is replaced by another wire of the same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?
The frequency of the stretched wire is given as:
Given that,
No.of harmonic n, length L, and tension T are the same in both cases.
Therefore,
Substituting the values of mass per unit length, we get
Solving the above the equation, we get v2 = 1/3 v1
Therefore, the frequency of the sitar is reduced by 1/2 of its actual value.
At what temperatures will the speed of sound in the air be 3 times its value at 0oC?
We know that,
Given that,
vT = 3v0
T = 2457 – 273 = 2184oC
When two waves of almost equal frequencies, n1 and n2, reach at a point simultaneously, what is the time interval between successive maxima?
It is given that the two frequencies are almost equal, that is n1 = n2.
For the formation of the beats, the frequencies must be n2 > n1.
The no.of frequencies per second in maxima = n = n2 – n1
Therefore, the time period of the maxima = 1/n = 1/n2 – n1 second.
Short AnswersA steel wire has a length of 12 m and a mass of 2.10 kg. What will be the speed of a transverse wave on this wire when a tension of 2.06 × 104 N is applied?
Length, l = 12 m
Total mass, M = 2.10 kg
m = M/l = 2.1/12
Tension, T = 2.06 × 104 N
Therefore, v is given as
v = 343.0 m/s
A pipe of 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz?
Given,
Length, l = 20 cm =0.2 m
v = 1237.5 Hz
Assuming that velocity of sound in air = 330 m/s
l = λ/4
λ = 4l
We know that,
v1= v/ λ = v/4l
v1 = 412.5 Hz
v/v1 = 1237.5/412.5 = 3/1
Therefore, the third harmonic will have 1237.5 Hz frequency.
A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz still air. The train beings to move at a speed of 10 m/s towards the platform. What is the frequency of the sound for an observer standing on the platform?
v0 = 400 Hz
vz = 10 m/s
The velocity of sound in air, va = 330 m/s
v’ is the frequency heard by the observer standing on the platform
v’ = (330)(400)/320 = 412.5 Hz
The wave pattern on a stretched string is shown in the figure. Interpret what kind of wave this is and find its wavelength.
When particles at t = T/4 and 3T/4 are at rest on a stationary wave which is at its mean position.
The nodes in the graph when the wave is at x = 10, 20, 30, 40 cm have a distance with the successive node is λ/2.
λ/2 = 30 – 20 = 20
λ = 20 cm
The pattern of standing waves formed on a stretched string at two instants of time is shown in the figure. The velocity of two waves superimposing to form stationary waves is 360 m/s, and their frequencies are 256 Hz.
a) calculate the time at which the second curve is plotted
b) mark nodes and antinodes on the curve
c) calculate the distance between A’ and C’
The frequency of the wave, v = 256 Hz
T = 1/v = 1/256 sec = 0.00390
T = 3.9 × 10-3 sec
a) When the particle’s mean position changes after T/4, the time is calculated as
t = T/4 = (3.9 10-3)/4 = 0.975 × 10-3 sec
t = 9.8 × 10-4 sec
b) Nodes: A, B, C, D, E
Antinode: A’, C’
c) The distance between A’ and C’ is
λ = v/f = 360/256 = 1.41 m
A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water. The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, the maximum intensity of sound is heard. If the room temperature is 20oC, calculate
a) speed of sound in air at room temperature
b) speed of sound in air at 0oC
c) if the water in the tube is replaced with mercury, will there be any difference in your observations?
Given,
The frequency of the tuning fork, f = 512 Hz
a) When the first maxima is considered, the length in the air column is, l = λ/4
λ = 4l
Speed of sound, v = f λ = 348.16 m/s
b) We know that,
