Question

the length of a simple pendulum increased by 60%. what will be the percentage change in the frequency of oscillations of the pendulum.

Please solve this question!!!!!​

Answer 1

Answer:

Let length of simple pendulum is L , time period is T

We know, time period is given by T = 2π√(L/g) -----(1)

A/C to question,

Length of simple pendulum = L + 300% of L

= L + 300L/100

= L + 3L

= 4L

Time period , T' = 2π√(4L/g) = 2 × 2π√(L/g)

From equation (1), T' = 2T

% change in time period = (T' - T)/T × 100

= (2T - T)/T × 100

Explanation:

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Answer 2

HEY MATE ✌️✌️

I HAVE ANSWER OF 44℅

IN THE ATTACHMENT IN PLACE OF 44℅ USE 60℅AND SOLVE IT

HOPE IT HELPS YOU✌️✌️