on suspending a body with a spring of force constant 100N/m its length increased by 2cm find the work done in this process
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time period of SHM is one spring and mass is connected spring constant ,k attached mass ,m so frequency is ,f = 2π √k/m as in both spring identical new spring constant for two parallel spring.
k new = k + k = 2 k time period of ascillation f new = 2 π √ n/k new = 2 π √ n/2k f new = 1/2 2π √n/k = f√2
Hence new Fequencey as f√ 2
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