The frequency n of vibrations of a string is given as n=(1/2l)√(T/m) where T is tension and l is the length of vibrating string. Then the dimensional formula for m is

Question

The frequency n of vibrations of a string is given as n=(1/2l)√(T/m) where T is tension and l is the length of vibrating string. Then the dimensional formula for m is (A) [M0L1T1] (B) [M0L0T0] (C) [M1L-1T0] (D) [M1L0T0]

Tardigrade Admin · Accepted Answer

n=(1/2l)√(T/m)⇒ n2=(1/4l2) (T/m) ⇒ m=(T/4l2n2)=[(MLT-2/L2× T-2)]=[ML-1]

Q. The frequency $n$ of vibrations of a string is given as $n = \frac{1}{2 l} \frac{T}{m}$ where $T$ is tension and $l$ is the length of vibrating string. Then the dimensional formula for $m$ is

$[M^{0} L^{1} T^{1}]$

$[M^{0} L^{0} T^{0}]$

$[M^{1} L^{- 1} T^{0}]$

$[M^{1} L^{0} T^{0}]$

$n = \frac{1}{2 l} \frac{T}{m} \Rightarrow n^{2} = \frac{1}{4 l ^{2}} \frac{T}{m}$
$\Rightarrow m = \frac{T}{4 l ^{2} n ^{2}} = [\frac{M L T ^{- 2}}{L ^{2} \times T ^{- 2}}] = [M L^{- 1}]$

Q. The frequency n of vibrations of a string is given as n=2l1​mT​​ where T is tension and l is the length of vibrating string. Then the dimensional formula for m is

[M0L1T1]

[M0L0T0]

[M1L−1T0]

[M1L0T0]