Q 1 :    

A closed and an open organ pipe have same lengths. If the ratio of frequencies of their seventh overtones is (a-1a) then the value of a is _________ .               [2024]



(16)      For closed organ pipe

             fc=(2n+1)v4l=15v4l

            For open organ pipe

           fo=(n+1)v2l=8v2l

           fcfo=1516=a-1aa=16

 



Q 2 :    

A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. The velocity of sound is _______ m/s.             [2024]



(294)       fc=v4l1    fo=v2l2

                v4×150-v2×350=7

                v600cm-v700cm=7

                v6m-v7m=7

                 v(142)=7

                  v=42×7=294m/s

 



Q 3 :    

In a closed organ pipe, the frequency of fundamental note is 30 Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz. If the organ pipe has a cross-sectional area of 2 cm2, the amount of water poured in the organ tube is ________ g. (Take speed of sound in air is 330 m/s)                     [2024]



(400)



Q 4 :    

A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is ________ Hz.                  [2024]



(6)          f=12LTμ

               f1=126μ,  f2=1254μ

               f1f2=13, f2-f1=12

                f1=6 Hz

 



Q 5 :    

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is 60 cm, the length of the closed pipe will be           [2024]

  • 30 cm

     

  • 45 cm

     

  • 60 cm

     

  • 15 cm

     

(4)

λ4=L1                 2(λ2)=λ=L2

f1=v4L1             f2=vL2

f1=f2

v4L1=vL2L2=4L1

60=4×L1L1=15 cm



Q 6 :    

A sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz when kept under some tension. The resonating length of the wire with a fundamental frequency of 600 Hz under the same tension is ____ cm.                  [2024]



(60)

f0=400 Hz;  v=Tμ=constant

λ2=L;  v=f0λ

v2f0=Lv=2Lf0

L'=v2f'=2Lf02f'

=Lf0f'=90×400600=60



Q 7 :    

Two open organ pipes of lengths 60 cm and 90 cm resonate at 6th and 5th harmonics respectively. The difference of frequencies for the given modes is ____ Hz

(Velocity of sound in air = 333 m/s)            [2024]



(740)

For first open organ pipe

f=nv2lf6=6v2l1

f6=6v2(0.6)=5v                 ...(i)

For second open organ pipe

f5=5v2(l2)=5v2(0.9)=259v               ...(ii)

Δf=f6-f5=5v-25v9

Δf=(45-25)v9=209v=209×333Δf=740 Hz