JEE MAIN - Physics (2003 - No. 45)
A metal wire of linear mass density of $$9.8$$ $$g/m$$ is stretched with a tension of $$10$$ $$kg$$-$$wt$$ between two rigid supports $$1$$ metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency $$n.$$ The frequency $$n$$ of the alternating source is
$$50$$ $$Hz$$
$$100$$ $$Hz$$
$$200$$ $$Hz$$
$$25$$ $$Hz$$
Explanation
KEY CONCEPT : For a string vibrating between two rigid support, the fundamental frequency is given by
$$n = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2 \times }}\sqrt {{{10 \times 9.8} \over {9.8 \times {{10}^{ - 3}}}}} = 50Hz$$
As the string is vibrating in resonance to a.c of frequency $$n,$$ therefore both the frequencies are same.
$$n = {1 \over {2\ell }}\sqrt {{T \over \mu }} = {1 \over {2 \times }}\sqrt {{{10 \times 9.8} \over {9.8 \times {{10}^{ - 3}}}}} = 50Hz$$
As the string is vibrating in resonance to a.c of frequency $$n,$$ therefore both the frequencies are same.
Comments (0)
