Waves Test

Result

Waves Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

A $$40\ cm$$ long brass rod is dropped, one end first on to a hard floor but it is caught before it topples over. With an oscilloscope it is determined that the impact produces a $$3\ kHz$$ tone. The speed of sound in brass is:

A
$$1200\ m/s$$

B
$$2400\ m/s$$

C
$$3600\ m/s$$

D
$$3000\ m/s$$

Solution

Both ends are free and therefore antinodes are formed.
The relation between the wavelength of the wave and the length of the rod for fundamental frequency will be:
$$\Rightarrow l=\dfrac{\lambda}{2} \ \ \Rightarrow \lambda=2l$$

The speed of the wave in the rod is:
$$v=f\lambda = 2\times 40\times 3\times 10^{3}$$$$=2400\ ms^{-1}$$
Question 2
1 / -0

When a sound wave of wavelength $$\lambda$$ is propagating in a medium the maximum velocity of the particle is equal to wave velocity. The amplitude of the wave is

A
$$\lambda$$

B
$${\dfrac{\lambda}{2}}$$

C
$$\displaystyle \dfrac{\lambda}{2\pi}$$

D
$$\displaystyle \dfrac{\lambda}{4\pi}$$

Solution

wave velocity of particle $$v=\dfrac{w}{K}$$
maximum particle velocity $$v_{2}=Aw$$
given $$Aw=\dfrac{w}{K}\ \ \ \ \ \ \Rightarrow A=\dfrac{1}{K}=\dfrac{1}{\dfrac{2\pi}{\lambda}}=\dfrac{\lambda}{2\pi}$$
Question 3
1 / -0

A wave travelling in positive X-direction with A = 0.2 m velocity = 360 m/s and $$\lambda$$= 60 m, then correct expression for the wave is : -

A
y = 0.2 sin $$\left [ 2\pi (6t+\frac{X}{60}) \right ]$$

B
y = 0.2 sin $$\left [\pi (6t+\frac{X}{60}) \right ]$$

C
y = 0.2 sin $$\left [ 2\pi (6t-\frac{X}{60}) \right ]$$

D
y = 0.2 sin $$\left [\pi (6t-\frac{X}{60}) \right ]$$

Solution
Question 4
1 / -0

A particle starts from a point $$P$$ at a distance of $$A/2$$ from the mean position $$O$$ & travels towards left as shown in the figure. If the time period of SHM, executed about $$O$$ is $$T$$ and amplitude $$A$$ then the equation of motion of particle is :

A
$$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{6}\right )$$

B
$$x = A \sin\left ( \dfrac{2\pi }{T} t+\dfrac{5\pi }{6}\right )$$

C
$$x = A \cos\left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{6}\right )$$

D
$$x = A \cos \left ( \dfrac{2\pi }{T} t+\dfrac{\pi }{3}\right )$$

Solution

$$\begin{array}{l} x=A\cos \, (\omega t+\theta ) \\ and\, \, \, v=\, \, -\, \omega A\sin \, \, (\omega t+\theta ) \\ \, \, at\, \, \, \, \, t=0, \\ \, x=\, A\, cos\theta \, \, \, \, \, and\, \, \, \, \, v=\, -\, \omega A\sin \theta \\ \, \, \, \, \, \therefore \, \, \, \, \dfrac { A }{ 2 } =A\, cos\theta \\ \Rightarrow \theta ={ \cos ^{ -1 } }\left( { \dfrac { 1 }{ 2 } } \right) =\dfrac { \pi }{ 3 } \\ now, \\ \, \, \, \, \, \, x=A\cos \, (\omega t+\frac { \pi }{ 3 } ) \\ \, \, \, \, \, \, \, \, \, \, =A\cos \, (\dfrac { { 2\pi } }{ T } .t+\dfrac { \pi }{ 3 } ) \\ \, \, \, \, \, \, \, \, \, \, =Asin\, (\dfrac { { 2\pi } }{ T } .t+\dfrac { \pi }{ 2 } +\dfrac { \pi }{ 3 } ) \\ \, \, \, \, \, \therefore \, \, \, A\, sin\, (\dfrac { { 2\pi } }{ T } .t+\dfrac { { 5\pi } }{ 6 } ) \\ so\, the\, correct\, option\, is\, \, B. \end{array}$$
Question 5
1 / -0

A small mass executes linear SHM about O with amplitude a and period T. Its displacement from O attime T/8 after passing through O is:

A
$$\dfrac{a}{8}$$

B
$$\dfrac{a}{2\sqrt{2}}$$

C
$$\dfrac{a}{2}$$

D
$$\dfrac{a}{\sqrt2}$$

Solution

Using the formula $$x=a\times cos(wt)$$ for SHM, and substituting $$t=\dfrac{T}{8 }$$and$$ w=\dfrac{w\times pi}{T}$$, we get $$x=a\times cos(\dfrac{pi}{4})=\dfrac{a}{ \sqrt(2)}$$
Question 6
1 / -0

Assertion :If Amplitude of SHM is doubled, the periodicity wall remain same.
Reason :Amplitude and periodicity are two independent characteristics of SHM.

A
Assertion & Reason are true and the reason explains the assertion

B
Assertion & Reason are true and the reason does not explain the assertion

C
Assertion is true. Reason is false

D
Both assertion and reason are false

Solution

Periodicity depends on other characteristics like length in case of pendulum, mass in case of spring mass system. But it is independent of amplitude.
Question 7
1 / -0

Statement-1 : Superposition principle is applicable only for small disturbances.
Statement-2 : Superposition principle is applicable only for non-linear waves.

A
If both the statements are true and statement-2 is the correct explanation of statement-1

B
If both the statements are true but statement-2 is not the correct explanation of statement-1

C
If statement-1 is true and statement-2 is false

D
If statement-1 is false and statement-2 is true

Solution

Superposition principle, states that, for all linear systems, the net response at a given place and time caused by two or more stimuli is the sum of the responses which would have been caused by each stimulus individually.
Question 8
1 / -0

The amplitude of the vibrating particle due to superposition of two $$SHMs$$,
$$y_{1}= \sin \left ( \omega t+\dfrac{\pi }{3} \right )$$ and $$y=2 \sin \omega t$$ is:

A
$$1$$

B
$$\sqrt{2}$$

C
$$\sqrt{7}$$

D
$$2$$

Solution

$$\begin{array}{l} { A^{ 2 } }=A_{ 1 }^{ 2 }+A_{ 2 }^{ 2 }+2{ A_{ 1 } }{ A_{ 2 } }\cos \emptyset \\ { A^{ 2 } }={ 1^{ 2 } }+{ 2^{ 2 } }+2\times 1\times 2\cos \left( { \pi /3 } \right) \\ { A^{ 2 } }=1+4+4\times \frac { 1 }{ 2 } \\ { A^{ 2 } }=5+2 \\ A=\sqrt { 7 } \end{array}$$
Question 9
1 / -0

When a wave pulse travelling in a string is reflected from a rigid wall to which string is tied as shown in figure. For this situation two statements are given below.
(1) The reflected pulse will be in same orientation of incident pulse due to a phase change of $$\pi$$ radians
(2) During reflection the wall exert a force on string in upward direction
For the above given two statements choose the correct option given below.

A
Only (1) is true

B
Only (2) is true

C
Both are true

D
Both are wrong

Solution

Reflected pulse will be inverted as it is reflected by a denser medium.
Since there will be phase change of $$\pi$$.
The wall exerts force in downward direction.
Question 10
1 / -0

Two identical pulses move in opposite directions with same uniform speeds on a stretched string. The width and kinetic energy of each pulse is $$L$$ and $$k$$ respectively. At the instant they completely overlap, the kinetic energy of the width $$L$$ of the string where they overlap is

A
$$k$$

B
$$2k$$

C
$$4k$$

D
$$8k$$

Solution

The velocity of the profile of each elementary section of the pulse is shown in figure 1 and figure 2.
When both the pulses completely overlap, the velocity profiles of both the pulses in overlap region are identical. By superposition, the velocity of each elementary section doubles. Therefore K.E. of each section becomes four times.Hence the K.E. in the complete width of overlap becomes four times, i.e., $$4k$$.