First slide

Standing waves

Question

A wire having a linear mass density 5.0 x $10^{- 3}$ kg/m is stretched between two rigid supports with a tension of 450 N. The wire resonates at a frequency of 420 Hz. The next higher frequency at which the same wire resonates is 480 Hz. The length of the wire is

Moderate

A

2.0 m
B

2.1 m
C

2.5 m
D

3 m

Solution

Two consecutive frequencies are 420 Hz and 480 Hz. So the fundamental frequency will be 60 Hz.
$∴ 60 = \frac{1}{2 \times l} \sqrt{\frac{450}{5 \times 10^{- 3}}} \Rightarrow l = 2.5 m$
Hence (2).

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App