1.Two adjacent piano keys are struck
simultaneously. The notes emitted by them have
frequencies \[n_{1}\] and \[n_{2}\] . The number of beats heard
per second is
a) \[\frac{1}{2}\left(n_{1}- n_{2}\right)\]
b) \[\frac{1}{2}\left(n_{1}+ n_{2}\right)\]
c) \[n_{1}\sim n_{2}\]
d) \[2\left(n_{1}- n_{2}\right)\]
Explanation: \[n_{1}\sim n_{2}\]
2. A tuning fork of frequency 100 when sounded
together with another tuning fork of unknown
frequency produces 2 beats per second. On
loading the tuning fork whose frequency is not
known and sounded together with a tuning fork of
frequency 100 produces one beat, then the
frequency of the other tuning fork is
a) 102
b) 98
c) 99
d) 101
Explanation:
3.A tuning fork sounded together with a tuning fork
of frequency 256 emits two beats. On loading the
tuning fork of frequency 256, the number of beats
heard are 1 per second. The frequency of tuning
fork is
a) 257
b) 258
c) 256
d) 254
Explanation:
4. If two tuning forks A and B are sounded together,
they produce 4 beats per second. A is then slightly
loaded with wax, they produce 2 beats when
sounded again. The frequency of A is 256. The
frequency of B will be
a) 250
b) 252
c) 260
d) 262
Explanation:
5. The frequencies of two sound sources are 256 Hz
and 260 Hz. At t = 0, the intensity of sound is
maximum. Then the phase difference at the time t
= 1/16 sec will be
a) Zero
b) \[\pi\]
c) \[\pi/2\]
d) \[\pi/4\]
Explanation: Time interval between two consecutive beats
6. Two tuning forks have frequencies 450 Hz and
454 Hz respectively. On sounding these forks
together, the time interval between successive
maximum intensities will be
a) 1/4 sec
b) 1/2 sec
c) 1 sec
d) 2 sec
Explanation:
7. When a tuning fork of frequency 341 is sounded
with another tuning fork, six beats per second are
heard. When the second tuning fork is loaded with
wax and sounded with the first tuning fork, the
number of beats is two per second. The natural
frequency of the second tuning fork is
a) 334
b) 339
c) 343
d) 347
Explanation:
8.Two tuning forks of frequencies 256 and 258
vibrations/sec are sounded together, then time
interval between consecutive maxima heard by
the observer is
a) 2 sec
b) 0.5 sec
c) 250 sec
d) 252 sec
Explanation:
9.A tuning fork gives 5 beats with another tuning
fork of frequency 100 Hz. When the first tuning
fork is loaded with wax, then the number of beats remains unchanged, then what will be the
frequency of the first tuning fork
a) 95 Hz
b) 100 Hz
c) 105 Hz
d) 110 Hz
Explanation:
10.Tuning fork \[F_{1}\] has a frequency of 256 Hz and it is
observed to produce 6 beats/second with another
tuning fork \[F_{2}\] . When \[F_{2}\] is loaded with wax, it
still produces 6 beats/second with \[F_{1}\] . The
frequency of \[F_{2}\] before loading was
a) 253 Hz
b) 262 Hz
c) 250 Hz
d) 259 Hz
Explanation:
