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1.

A string is stretched by a force of 40 newton. The mass of 10 m length of this string is 0.01 kg. the speed of transverse waves in this string will beA. 400m/sB. 40 m/sC. 200 m/sD. 80 m/s

Answer» Correct Answer - C
`V=sqrt(T/(mu))`
`=sqrt((40xx10)/0.01)=200 m//s`
Discussion
2.

Graph shows three waves that are separately sent along a string that is stretched under a certain tension along x-axis. If `omega_(1),omega_(2) and omega_(3)` are their angular frequencies, respectively, then: A. `omega_(1)=omega_(3) gt omega_(2)`B. `omega_(1)gtomega_(2)gtomega_(3)`C. `omega_(2)gtomega_(1)=omega_(3)`D. `omega_(1)=omega_(2)=omega_(3)`

Answer» Correct Answer - A
(A) As `v=sqrt(T/(mu))` is same for all, wave with maximum wavelength will have minimum angular frequency (by `v=n lambda`). Also as `lambda_(1)=lambda_(2)` thus `omega_(1)=omega_(2)` hence (A)
Discussion
3.

Two interferring waves have the same wavelength, frequency and amplitude. They are travelling in the same direction but `90^(@)` out of phase compared to individual waves. The resultant wave will have the same.A. amplitude and velocity but different wavelengthB. frequency and velocity but different wavelengthC. wavelength and velocity but different amplitudeD. amplitude and frequency but different wavelength

Answer» Correct Answer - C
Wavelength and velocity are medium dependent. Final amplitude is decided by the superposition of individual amplitudes.
Discussion
4.

Sinusoidal waves `5.00 cm ` in amplitude are to be transmitted along a string having a linear mass density equal to `4.00xx10^-2kg//m`. If the source can deliver a maximum power of `90W` and the string is under a tension of `100N`, then the highest frequency at which the source can operate is (take `pi^2=10`)A. 45.3 HzB. 50 HzC. 30 HzD. 62.3 Hz

Answer» Correct Answer - C
`(C) P=1/2 mu omega^(2) A^(2)v` using `V=sqrt(T/(mu))`
`P=1/2 omega^(2)A^(2)sqrt(Tmu) rArr omega=sqrt((2p)/(A^(2)sqrt(Tmu))) f=(omega)/(2pi)=1/(2pi)sqrt((2P)/(A^(2)sqrt(Tmu)))`
using data f=30 Hz
Discussion
5.

The displacement from the position of equilibrium of a point 4 cm from a source of sinusoidal oscillations is half the amplitude at the moment `t=T//6` (T is the time perios). Assume that the source was at mean position at `t=0`. The wavelength of the running wave is :A. 0.96 mB. 0.48 mC. 0.24 mD. 0.12 m

Answer» Correct Answer - B
Equation of wave is `y=A sin (omegat-kx)`
`rArr A/2=A sin ((2pi)/TxxT/6-(2pi)/(lambda)xx4)`
`rArr (pi)/6=(pi)/3-(2pi)/(lambda)xx4rArr (2pi)/(lambda)xx4=(pi)/6`
`rArr lambda=48 cm`
`rArr lambda=0.48 meters`
Discussion
6.

The equetion of a wave travelling on a string is `y = 4 sin(pi)/(2)(8t-(x)/(8))` if x and y are in centimetres, then velocity of waves isA. 64 cm/s in -x directionB. 32 cm/s in -x directionC. 32 cm/s in +x directionD. 64 cm/s in +x direction

Answer» Correct Answer - D
direction positive
`V=(omega)/K =64 cm//sec`
Discussion
7.

`S_(1)` : A standing wave pattern if formed in a string. The power transfer through a point (other than node and antinode) is zero always `S_(2)`: if the equation of transverse wave is `y= sin 2pi[t/0.04-x/40]`, where distance is in cm. time in second, then the wavelength will be 40 cm. `S_(3)`: if the length of the vibrating string is kept constant, then frequency of the string will be directly proportional to `sqrt(T)`A. FTTB. TTFC. TFTD. FFF

Answer» Correct Answer - A
`S_(1)=` False
at node v=0, at antinode tension `bot` to velocity `:.` at the points power `=0(P=vec(F).vec(V))`
At other points `Pne0`
`S_(2)`: True
Equation of transverse wave is given by
`y=5 sin 2pi[t/0.04-x/40].....(1)`
The standard equation of transverse wave is
`y=a sin 2pi[t/T-x/(lambda)].....(2)`
Now comparing the given equation (1) and the standard equation (2) we obtain `lambda=40 cm ` where `lambda` is the wavelength of wave
`S_(3)`: True
The effect of tension is searched by keeping the length of the string constant, while the tension is alternate to brign the wire into unision with a number of standerd forks in turn. experiments varifies that the frequency is directly proportional to the square root of tension which is
`f prop sqrt(T)`
Discussion
8.

Assertion: In a small segment of string carrying sinusoidal wave, total energy is conserved. Reason: Every small part moves in SHM and total energy of SHM is conserved.A. if both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and reason are true, but reason is not correct explanation of the Assertion.C. If Assertion is true, but the reason is falseD. If assertion is false, but the reason is true.

Answer» Correct Answer - D
Every small segment is acted upon by forces from both sides of it hence energy is not conserved, rather it is transmitted by the element.
Discussion
9.

The equation of a progressive wave is given by `y=a sin (628t-31.4x)`. If the distance are expressed in cms and time seconds, then the wave in this string wll beA. 314 cnB. 628 cmC. 5 cmD. 400 cm

Answer» Correct Answer - C
`k=(2pi)/(lambda)=31.4`
`lambda=5 cm`
Discussion
10.

Two small boats are `10m` apart on a lake. Each pops up and down with a period of 4.0 seconds due to wave motion on the surface of water. When one boat is at its highest point, the other boat is its lowest point. Both boats are always within a single cycle of the waves. The speed of the waves is:A. 2m/secB. 2.5 m/sC. 10 m/sD. 5 m/sec

Answer» Correct Answer - D
distance between crest and trough `=(lambda)/2=10 m`
`rArr lambda=20 m`
time period T=4sec
`:. V=lambda//T =20 m //4 sec`
=5 m/s
Discussion
11.

two particle of medium disturbed by the wave propagation are at `x_(1)=0 and x_(2)=1 cm`. The respective displacement (in cm) of the particles can be given by the equation: `y_(1)=2 sin 3pi t, y_(2) sin (3pi t-pi//8)` the wave velocity isA. 16 cm/secB. 24 cm/secC. 12 cm/secD. 8 cm/sec

Answer» Correct Answer - B
Given `omega=3pi`
`:. f=(omega)/(2pi)=1.5`,
Also `Deltax=1.0 cm`
Given `phi=(2pi)/(lambda) DeltaxrArr (pi)/8=(2pi)/(lambda)xx1`
`rArr lambda=16 cm " " v= f lambda=16xx1.5 =24 cm //sec`
Discussion