A)
410
cycles/sec
B)
500 cycles/sec
C)
550
cycles/sec
D)
450 cycles/sec
Solution:
.
. The displacement of a particle is represented by
the equation .This
motion of particle is
A)
Simple
harmonic with period
B)
Simple
harmonic with period
C)
Periodic but not
simple harmonic.
D)
Non- periodic
Solution:
As
given that, y =
Velocity
of the particle
So, acceleration,
Hence, due to negative
sign motion is simple harmonic motion (SHM.)
A simple harmonic motion
is always periodic. So motion is periodic simple harmonic.
From the given equation,
Compare it
by standard equation
So,
Hence, the motion is
SHM with period.
3. The displacement of a particle is
represented by the equation .This
motion is
A)
Non- periodic
B)
Periodic but not simple
harmonic
C) Simple harmonic with
period
D)
Simple
harmonic with period
Solution:
A motion will be
harmonic if a α displacement and a simple harmonic motion is always
periodic but all simple harmonic motion are periodic but all periodic are not
harmonic. As given equation of motion is
Again, differentiating
both sides w.r.t ‘t’
So, is not directly proportional to y
So motion is not
harmonic.
4. The
relation between acceleration and displacement of four particles are given
below
A)
B)
C)
D)
Solution:
For motion to be SHM
acceleration of the particle must be opposite of restoring force and
proportional to negative of displacement. So,
i.e., ,
so Hence,
We should be clear that has to be linear.
(a)
98 cm
(b) 140 cm
(c) 120 cm
(d) 144 cm
Solution:
=98cm
A)
600
cps
B)
660 cps
C)
990
cps
D)
330 cps
Solution:
A)
166 m/sec
B)
664 m/sec
C)
332
m/sec
D)
1328 m/sec
Solution:
.
A)
384
B)
192
C)
300
D)
200
Solution:
Wave number = but = and
A)
8 m/s
(approx.)
B)
800 m/s
C)
7
m/s
D)
6 m/s (approx.)
Solution:
Since, source is moving towards the
listener so
A)
900 Hz
B)
625 Hz
C)
750
Hz
D)
800 Hz
Solution:
A)
B)
v
C)
D)
3v
Solution:
A)
10/9
B)
11/10
C)
(11/10)2
A)
600
B)
1050
C)
1400
D)
2400
Solution:
= 1400
cps
A)
600
Hz
B)
1200 Hz
C)
1500
Hz
D)
1600 Hz
Solution:
A)
225
Hz
B)
200 Hz
C)
150
Hz
D)
100 Hz
Solution:
A)
9 :
8
B)
8 : 9
C)
1 :
1
D)
9 : 10
Solution:
When source is approaching the observer,
the frequency heard
When source is receding, the frequency
heard
A)
750
Hz
B)
857 Hz
C)
1143
Hz
D)
1333 Hz
Solution:
By using
A)
1000
Hz
B)
1066 Hz
C)
941
Hz
D)
352 Hz
Solution:
For source S =0.70×2π×5 =22 m/sec
Minimum frequency is heard when the source is receding the man. It is
given by
A)
8
B)
6
C)
4
D)
2
Solution:
For direct
sound source is moving away from the observes so frequency heard in this
case
The other
sound is echo, reaching the observer from the wall and can be regarded as
coming from the image of source formed by reflection at the wall. This image is
approaching the observer in the direction of sound.
Hence for
reflected sound, frequency heard by the observer is
A)
109 vib/sec
B)
132 vib/sec
C)
140 vib/sec
D)
248 vib/sec
Solution:
= 140 vib/sec