Length of a stretched wire is 2m. It is oscillating in its fourth overtone mode. Maximum amplitude of oscillations is 2mm. Find amplitude of oscillation at a distance of 0.2m from one fixed end.

Amplitude of oscillating field is 2 xx 10^-6 T . The amplitude of other oscillating field is

A string of length L is fixed at both ends . It is vibrating in its 3rd overtone with maximum amplitude a. The amplitude at a distance L//3 from one end is

An object of mass M oscillating with amplitude 'A' mass 'm' is added at mean position. Find new amplitude of oscillation.

A string of length 'L' is fixed at both ends . It is vibrating in its 3rd overtone with maximum amplitude 'a'. The amplitude at a distance L//3 from one end is

What is the amplitude of damped oscillator if time becomes t= 2m ?

A string of length 'l' is fixed at both ends. It is vibrating in its 3^(rd) overtone with maximum ampltiude 'a' . The amplitude at a distance (l)/(3) from one end is = sqrt(p)(a)/(2) . Find p .

When an oscillator completes 50 oscillations its amplitude reduced to half of initial value (A_0) . The amplitude of oscillation, when it completes 150 oscillations is

An iron ring of mass m_1 oscillates in its an plane with time period of pi sec. Assume that the amplitude of oscillation is small. When it is passing through the mean position, a magnet of mass m is attached to the lowest point. If the amplitude of oscillation decreases by factor of 7, find m/(m_1) .