â ‹explain this Q The wire shown in the figure is vibrating in the fundamental mode - Physics -

Question

â€‹explain this

Q The wire shown in the figure is vibrating in the
fundamental mode with a frequency 750 Hz The suspended mass (M) is
submerged in a liquid whose density is 5/9 times that of the
mass
L = 1 m, M = 9 kg

Aarushi Mishra · Accepted Answer

Dear studentI am assuming that the question says "
density of liquid is 59 times mass density"L=1mM=9kgf=750 HzLet the volume of mass M be VDensity of liquid, ρ=59Mass density=59×MV kg m-3=59×9V kg m-3=5V kg m-3 Since density=massvolumeWe know that frequency, f=n2LTμI first case mass is simply suspended so tension in the string, T=Mg=9g750=n2L9gμ __________1Now the mass is submerged in liquid of density ρ
<img alt="" src=
"https://s3mn%20mnimgs%20com/img/shared/content_ck_images/ck_5a9b6d4cbd3e5%20png"
style="opacity: 0 9;">

FB buoyant forceFB=ρVg=5V×Vg=5gMg=9gSince M is suspended it will be in equilibriumT'+FB=MgT'=Mg-FB=9g-5g=4gNew frequencyf'=n2LT'μNote, n will remain same,  frequency will change due to tension onlyf'=n2L4gμ__________________2Dividing equation 1 and 2750f'=n2L9gμn2L4gμ750f'=94=32f'=23×750f'=500Hz

​explain this Q. The wire shown in the figure is vibrating in the fundamental mode with a frequency 750 Hz. The suspended mass (M) is submerged in a liquid whose density is 5/9 times that of the mass. L = 1 m, M = 9 kg.

explain this

Q. The wire shown in the figure is vibrating in the fundamental mode with a frequency 750 Hz. The suspended mass (M) is submerged in a liquid whose density is 5/9 times that of the mass.
L = 1 m, M = 9 kg.