wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A steel wire of length 116 cm is connected to an aluminum wire of length 150 cm and stretched between two fixed supports. The tension produced is 52 N. The cross-section of each wire is 2 mm2. If a transverse wave is set up in the wire, find the lowest frequency for which standing waves with a node at the joint are produced. (density of aluminum (ρal)=2.6 g/cm3 and density of steel (ρs)=7.8 g/cm3).


A
50 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
100 Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
150 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
200 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 100 Hz
Given that
ρal=2.6 g/cm3, ρs=7.8 g/cm3
Aa=As=2 mm2=2×106 m2
Tension, T=52 N
Length of aluminium wire, Lal=150 cm
Length of steel wire, Ls=116 cm
Let μal be the linear mass density of aluminium, and μs be the linear mass density for steel.

So, μal=ρal×Aal=(2.6×103 kg/m3)×(2×106 m2)=5.2×103 kg/m
and μs=ρs×As=(7.8×103 kg/m3)×(2×106 m2)=15.6×103 kg/m

Now for wave speed,
val=Tμal=525.2×103=100 m/s
and
vs=Tμs=5215.6×10358 m/s

For node at the joint, let us consider number of loops in aluminium is nal and in steel is ns.
For lowest frequency, fal=fs
nalval2Lal=nsvs2Ls
nalns=(vsval)×(LalLs)=58100×150116=34

Therefore, for minimum frequency,
fal=3×val2Lal=3×1002×150×102=100 Hz

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon